MAYBE
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<BR><B>Left Termination</B> of the query pattern
p_in_1(a)
w.r.t. the given <I>Prolog program</I> could not be shown:<BR><BR><BR><BR><pre>&#8627 <B>Prolog</B></pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>  &#8627 PrologToPiTRSProof</pre><BR>Clauses:<BR><BR>p(X)&#160;:-&#160;','(l(X), q(X)).<BR>q(.(A, [])).<BR>r(1).<BR>l([]).<BR>l(.(H, T))&#160;:-&#160;','(r(H), l(T)).<BR><BR>Queries:<BR><BR>p(a).<BR><BR>We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
<BR><FONT COLOR=#0000cc>p_in</font>: (f)
<BR><FONT COLOR=#0000cc>l_in</font>: (f)
<BR>Transforming <I>Prolog</I> into the following <B>Term Rewriting System</B>:
<BR>Pi-finite rewrite system:<BR>The TRS R consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>p_in_a</font>(<FONT COLOR=#cc0000>X</font>) &#8594; <FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#0000cc>[]</font>) &#8594; <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#0000cc>[]</font>)
<BR><FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>H</font>))
<BR><FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#0000cc>1</font>) &#8594; <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)
<BR><FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>T</font>))
<BR><FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>))
<BR><FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>A</font>, <FONT COLOR=#0000cc>[]</font>)) &#8594; <FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>A</font>, <FONT COLOR=#0000cc>[]</font>))
<BR><FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>X</font>)</BLOCKQUOTE><BR>The argument filtering Pi contains the following mapping:<BR><FONT COLOR=#0000cc>p_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>p_in_a</font><BR>
<FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>l_in_a</font><BR>
<FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>r_in_a</font><BR>
<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>[]</font>&#160; = &#160;<FONT COLOR=#0000cc>[]</font><BR>
<FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>q_out_g</font><BR>
<FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<BR>
<P><B>Infinitary Constructor Rewriting Termination</B> of PiTRS implies <B>Termination</B> of Prolog<P>
<BR><BR><pre>&#8627 Prolog</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>    &#8627 <B>PiTRS</B></pre><pre>      &#8627 DependencyPairsProof</pre><pre>  &#8627 PrologToPiTRSProof</pre><BR>Pi-finite rewrite system:<BR>The TRS R consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>p_in_a</font>(<FONT COLOR=#cc0000>X</font>) &#8594; <FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#0000cc>[]</font>) &#8594; <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#0000cc>[]</font>)
<BR><FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>H</font>))
<BR><FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#0000cc>1</font>) &#8594; <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)
<BR><FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>T</font>))
<BR><FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>))
<BR><FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>A</font>, <FONT COLOR=#0000cc>[]</font>)) &#8594; <FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>A</font>, <FONT COLOR=#0000cc>[]</font>))
<BR><FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>X</font>)</BLOCKQUOTE><BR>The argument filtering Pi contains the following mapping:<BR><FONT COLOR=#0000cc>p_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>p_in_a</font><BR>
<FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>l_in_a</font><BR>
<FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>r_in_a</font><BR>
<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>[]</font>&#160; = &#160;<FONT COLOR=#0000cc>[]</font><BR>
<FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>q_out_g</font><BR>
<FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<BR><BR>Using Dependency Pairs [1,30] we result in the following initial DP problem:<BR>Pi DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>P_IN_A</font>(<FONT COLOR=#cc0000>X</font>) &#8594; <FONT COLOR=#0000cc>U1_A</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>P_IN_A</font>(<FONT COLOR=#cc0000>X</font>) &#8594; <FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#cc0000>X</font>)
<BR><FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>H</font>))
<BR><FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>R_IN_A</font>(<FONT COLOR=#cc0000>H</font>)
<BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>U4_A</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>T</font>))
<BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#cc0000>T</font>)
<BR><FONT COLOR=#0000cc>U1_A</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>U2_A</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>U1_A</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>Q_IN_G</font>(<FONT COLOR=#cc0000>X</font>)</BLOCKQUOTE><BR>The TRS R consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>p_in_a</font>(<FONT COLOR=#cc0000>X</font>) &#8594; <FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#0000cc>[]</font>) &#8594; <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#0000cc>[]</font>)
<BR><FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>H</font>))
<BR><FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#0000cc>1</font>) &#8594; <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)
<BR><FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>T</font>))
<BR><FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>))
<BR><FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>A</font>, <FONT COLOR=#0000cc>[]</font>)) &#8594; <FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>A</font>, <FONT COLOR=#0000cc>[]</font>))
<BR><FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>X</font>)</BLOCKQUOTE><BR>The argument filtering Pi contains the following mapping:<BR><FONT COLOR=#0000cc>p_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>p_in_a</font><BR>
<FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>l_in_a</font><BR>
<FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>r_in_a</font><BR>
<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>[]</font>&#160; = &#160;<FONT COLOR=#0000cc>[]</font><BR>
<FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>q_out_g</font><BR>
<FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>R_IN_A</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>R_IN_A</font><BR>
<FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>U1_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U1_A</font>(<FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>U4_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U4_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>U2_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U2_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>L_IN_A</font><BR>
<FONT COLOR=#0000cc>P_IN_A</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>P_IN_A</font><BR>
<FONT COLOR=#0000cc>Q_IN_G</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>Q_IN_G</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<BR>We have to consider all (P,R,Pi)-chains<BR><BR><pre>&#8627 Prolog</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>    &#8627 PiTRS</pre><pre>      &#8627 DependencyPairsProof</pre><pre>        &#8627 <B>PiDP</B></pre><pre>          &#8627 DependencyGraphProof</pre><pre>  &#8627 PrologToPiTRSProof</pre><BR>Pi DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>P_IN_A</font>(<FONT COLOR=#cc0000>X</font>) &#8594; <FONT COLOR=#0000cc>U1_A</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>P_IN_A</font>(<FONT COLOR=#cc0000>X</font>) &#8594; <FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#cc0000>X</font>)
<BR><FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>H</font>))
<BR><FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>R_IN_A</font>(<FONT COLOR=#cc0000>H</font>)
<BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>U4_A</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>T</font>))
<BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#cc0000>T</font>)
<BR><FONT COLOR=#0000cc>U1_A</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>U2_A</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>U1_A</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>Q_IN_G</font>(<FONT COLOR=#cc0000>X</font>)</BLOCKQUOTE><BR>The TRS R consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>p_in_a</font>(<FONT COLOR=#cc0000>X</font>) &#8594; <FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#0000cc>[]</font>) &#8594; <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#0000cc>[]</font>)
<BR><FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>H</font>))
<BR><FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#0000cc>1</font>) &#8594; <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)
<BR><FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>T</font>))
<BR><FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>))
<BR><FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>A</font>, <FONT COLOR=#0000cc>[]</font>)) &#8594; <FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>A</font>, <FONT COLOR=#0000cc>[]</font>))
<BR><FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>X</font>)</BLOCKQUOTE><BR>The argument filtering Pi contains the following mapping:<BR><FONT COLOR=#0000cc>p_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>p_in_a</font><BR>
<FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>l_in_a</font><BR>
<FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>r_in_a</font><BR>
<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>[]</font>&#160; = &#160;<FONT COLOR=#0000cc>[]</font><BR>
<FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>q_out_g</font><BR>
<FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>R_IN_A</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>R_IN_A</font><BR>
<FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>U1_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U1_A</font>(<FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>U4_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U4_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>U2_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U2_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>L_IN_A</font><BR>
<FONT COLOR=#0000cc>P_IN_A</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>P_IN_A</font><BR>
<FONT COLOR=#0000cc>Q_IN_G</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>Q_IN_G</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<BR>We have to consider all (P,R,Pi)-chains<BR>The approximation of the Dependency Graph [30] contains 1 SCC with 6 less nodes.<BR><BR><pre>&#8627 Prolog</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>    &#8627 PiTRS</pre><pre>      &#8627 DependencyPairsProof</pre><pre>        &#8627 PiDP</pre><pre>          &#8627 DependencyGraphProof</pre><pre>            &#8627 <B>PiDP</B></pre><pre>              &#8627 UsableRulesProof</pre><pre>  &#8627 PrologToPiTRSProof</pre><BR>Pi DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#cc0000>T</font>)
<BR><FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>H</font>))</BLOCKQUOTE><BR>The TRS R consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>p_in_a</font>(<FONT COLOR=#cc0000>X</font>) &#8594; <FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#0000cc>[]</font>) &#8594; <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#0000cc>[]</font>)
<BR><FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>H</font>))
<BR><FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#0000cc>1</font>) &#8594; <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)
<BR><FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>T</font>))
<BR><FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>))
<BR><FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>A</font>, <FONT COLOR=#0000cc>[]</font>)) &#8594; <FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>A</font>, <FONT COLOR=#0000cc>[]</font>))
<BR><FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>X</font>)</BLOCKQUOTE><BR>The argument filtering Pi contains the following mapping:<BR><FONT COLOR=#0000cc>p_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>p_in_a</font><BR>
<FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>l_in_a</font><BR>
<FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>r_in_a</font><BR>
<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>[]</font>&#160; = &#160;<FONT COLOR=#0000cc>[]</font><BR>
<FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>q_out_g</font><BR>
<FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>L_IN_A</font><BR>
<BR>We have to consider all (P,R,Pi)-chains<BR>For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.<BR><BR><pre>&#8627 Prolog</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>    &#8627 PiTRS</pre><pre>      &#8627 DependencyPairsProof</pre><pre>        &#8627 PiDP</pre><pre>          &#8627 DependencyGraphProof</pre><pre>            &#8627 PiDP</pre><pre>              &#8627 UsableRulesProof</pre><pre>                &#8627 <B>PiDP</B></pre><pre>                  &#8627 PiDPToQDPProof</pre><pre>  &#8627 PrologToPiTRSProof</pre><BR>Pi DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#cc0000>T</font>)
<BR><FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>H</font>))</BLOCKQUOTE><BR>The TRS R consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#0000cc>1</font>) &#8594; <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)</BLOCKQUOTE><BR>The argument filtering Pi contains the following mapping:<BR><FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>r_in_a</font><BR>
<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>L_IN_A</font><BR>
<BR>We have to consider all (P,R,Pi)-chains<BR>Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.<BR><BR><pre>&#8627 Prolog</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>    &#8627 PiTRS</pre><pre>      &#8627 DependencyPairsProof</pre><pre>        &#8627 PiDP</pre><pre>          &#8627 DependencyGraphProof</pre><pre>            &#8627 PiDP</pre><pre>              &#8627 UsableRulesProof</pre><pre>                &#8627 PiDP</pre><pre>                  &#8627 PiDPToQDPProof</pre><pre>                    &#8627 <B>QDP</B></pre><pre>                      &#8627 Rewriting</pre><pre>  &#8627 PrologToPiTRSProof</pre><BR>Q DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>L_IN_A</font> &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_in_a</font>)
<BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font></BLOCKQUOTE><BR>The TRS R consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>r_in_a</font> &#8594; <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)</BLOCKQUOTE><BR>The set Q consists of the following terms:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>r_in_a</font></BLOCKQUOTE><BR>We have to consider all (P,Q,R)-chains.<BR>By rewriting [15] the rule <FONT COLOR=#0000cc>L_IN_A</font> &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_in_a</font>) at position [0] we obtained the following new rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>L_IN_A</font> &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>))</BLOCKQUOTE><BR><BR><BR><pre>&#8627 Prolog</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>    &#8627 PiTRS</pre><pre>      &#8627 DependencyPairsProof</pre><pre>        &#8627 PiDP</pre><pre>          &#8627 DependencyGraphProof</pre><pre>            &#8627 PiDP</pre><pre>              &#8627 UsableRulesProof</pre><pre>                &#8627 PiDP</pre><pre>                  &#8627 PiDPToQDPProof</pre><pre>                    &#8627 QDP</pre><pre>                      &#8627 Rewriting</pre><pre>                        &#8627 <B>QDP</B></pre><pre>                          &#8627 UsableRulesProof</pre><pre>  &#8627 PrologToPiTRSProof</pre><BR>Q DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>L_IN_A</font> &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>))
<BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font></BLOCKQUOTE><BR>The TRS R consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>r_in_a</font> &#8594; <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)</BLOCKQUOTE><BR>The set Q consists of the following terms:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>r_in_a</font></BLOCKQUOTE><BR>We have to consider all (P,Q,R)-chains.<BR>As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.<BR><BR><pre>&#8627 Prolog</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>    &#8627 PiTRS</pre><pre>      &#8627 DependencyPairsProof</pre><pre>        &#8627 PiDP</pre><pre>          &#8627 DependencyGraphProof</pre><pre>            &#8627 PiDP</pre><pre>              &#8627 UsableRulesProof</pre><pre>                &#8627 PiDP</pre><pre>                  &#8627 PiDPToQDPProof</pre><pre>                    &#8627 QDP</pre><pre>                      &#8627 Rewriting</pre><pre>                        &#8627 QDP</pre><pre>                          &#8627 UsableRulesProof</pre><pre>                            &#8627 <B>QDP</B></pre><pre>                              &#8627 QReductionProof</pre><pre>  &#8627 PrologToPiTRSProof</pre><BR>Q DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>L_IN_A</font> &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>))
<BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font></BLOCKQUOTE><BR>R is empty.<BR>The set Q consists of the following terms:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>r_in_a</font></BLOCKQUOTE><BR>We have to consider all (P,Q,R)-chains.<BR>We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>r_in_a</font></BLOCKQUOTE><BR><BR><BR><pre>&#8627 Prolog</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>    &#8627 PiTRS</pre><pre>      &#8627 DependencyPairsProof</pre><pre>        &#8627 PiDP</pre><pre>          &#8627 DependencyGraphProof</pre><pre>            &#8627 PiDP</pre><pre>              &#8627 UsableRulesProof</pre><pre>                &#8627 PiDP</pre><pre>                  &#8627 PiDPToQDPProof</pre><pre>                    &#8627 QDP</pre><pre>                      &#8627 Rewriting</pre><pre>                        &#8627 QDP</pre><pre>                          &#8627 UsableRulesProof</pre><pre>                            &#8627 QDP</pre><pre>                              &#8627 QReductionProof</pre><pre>                                &#8627 <B>QDP</B></pre><pre>                                  &#8627 Instantiation</pre><pre>  &#8627 PrologToPiTRSProof</pre><BR>Q DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>L_IN_A</font> &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>))
<BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font></BLOCKQUOTE><BR>R is empty.<BR>Q is empty.<BR>We have to consider all (P,Q,R)-chains.<BR>By instantiating [15] the rule <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font> we obtained the following new rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font></BLOCKQUOTE><BR><BR><BR><pre>&#8627 Prolog</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>    &#8627 PiTRS</pre><pre>      &#8627 DependencyPairsProof</pre><pre>        &#8627 PiDP</pre><pre>          &#8627 DependencyGraphProof</pre><pre>            &#8627 PiDP</pre><pre>              &#8627 UsableRulesProof</pre><pre>                &#8627 PiDP</pre><pre>                  &#8627 PiDPToQDPProof</pre><pre>                    &#8627 QDP</pre><pre>                      &#8627 Rewriting</pre><pre>                        &#8627 QDP</pre><pre>                          &#8627 UsableRulesProof</pre><pre>                            &#8627 QDP</pre><pre>                              &#8627 QReductionProof</pre><pre>                                &#8627 QDP</pre><pre>                                  &#8627 Instantiation</pre><pre>                                    &#8627 <B>QDP</B></pre><pre>                                      &#8627 NonTerminationProof</pre><pre>  &#8627 PrologToPiTRSProof</pre><BR>Q DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>L_IN_A</font> &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>))
<BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font></BLOCKQUOTE><BR>R is empty.<BR>Q is empty.<BR>We have to consider all (P,Q,R)-chains.<BR>We used the non-termination processor [17] to show that the DP problem is infinite.<BR>Found a loop by narrowing to the left:<BR><BR>The TRS P consists of the following rules:<BLOCKQUOTE><BR><FONT COLOR=#0000cc>L_IN_A</font> &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>))
<BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font></BLOCKQUOTE><BR>The TRS R consists of the following rules:none<BR><BR><BR>s = <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)) evaluates to  t =<FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>))<BR><BR>Thus s starts an infinite chain as s semiunifies with t with the following substitutions:<BR><UL><LI> Semiunifier: [ ]</LI>
<LI> Matcher: [ ]</LI></UL><BR><BR><hr><BR><B>Rewriting sequence</B><BR><BR><B><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>))</B> &#8594; <FONT COLOR=#0000cc>L_IN_A</font><BR>with rule <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font> at position [] and matcher [ ]<BR><BR><B><FONT COLOR=#0000cc>L_IN_A</font></B> &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>))<BR>with rule <FONT COLOR=#0000cc>L_IN_A</font> &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>))<BR><BR>Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence<BR><BR><BR>All these steps are and every following step will be a correct step w.r.t to Q.<BR><BR><BR><BR><BR>We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
<BR><FONT COLOR=#0000cc>p_in</font>: (f)
<BR><FONT COLOR=#0000cc>l_in</font>: (f)
<BR>Transforming <I>Prolog</I> into the following <B>Term Rewriting System</B>:
<BR>Pi-finite rewrite system:<BR>The TRS R consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>p_in_a</font>(<FONT COLOR=#cc0000>X</font>) &#8594; <FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#0000cc>[]</font>) &#8594; <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#0000cc>[]</font>)
<BR><FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>H</font>))
<BR><FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#0000cc>1</font>) &#8594; <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)
<BR><FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>T</font>))
<BR><FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>))
<BR><FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>A</font>, <FONT COLOR=#0000cc>[]</font>)) &#8594; <FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>A</font>, <FONT COLOR=#0000cc>[]</font>))
<BR><FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>X</font>)</BLOCKQUOTE><BR>The argument filtering Pi contains the following mapping:<BR><FONT COLOR=#0000cc>p_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>p_in_a</font><BR>
<FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>l_in_a</font><BR>
<FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>r_in_a</font><BR>
<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>[]</font>&#160; = &#160;<FONT COLOR=#0000cc>[]</font><BR>
<FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<BR>
<P><B>Infinitary Constructor Rewriting Termination</B> of PiTRS implies <B>Termination</B> of Prolog<P>
<BR><BR><pre>&#8627 Prolog</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>    &#8627 <B>PiTRS</B></pre><pre>      &#8627 DependencyPairsProof</pre><BR>Pi-finite rewrite system:<BR>The TRS R consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>p_in_a</font>(<FONT COLOR=#cc0000>X</font>) &#8594; <FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#0000cc>[]</font>) &#8594; <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#0000cc>[]</font>)
<BR><FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>H</font>))
<BR><FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#0000cc>1</font>) &#8594; <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)
<BR><FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>T</font>))
<BR><FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>))
<BR><FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>A</font>, <FONT COLOR=#0000cc>[]</font>)) &#8594; <FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>A</font>, <FONT COLOR=#0000cc>[]</font>))
<BR><FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>X</font>)</BLOCKQUOTE><BR>The argument filtering Pi contains the following mapping:<BR><FONT COLOR=#0000cc>p_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>p_in_a</font><BR>
<FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>l_in_a</font><BR>
<FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>r_in_a</font><BR>
<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>[]</font>&#160; = &#160;<FONT COLOR=#0000cc>[]</font><BR>
<FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<BR><BR>Using Dependency Pairs [1,30] we result in the following initial DP problem:<BR>Pi DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>P_IN_A</font>(<FONT COLOR=#cc0000>X</font>) &#8594; <FONT COLOR=#0000cc>U1_A</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>P_IN_A</font>(<FONT COLOR=#cc0000>X</font>) &#8594; <FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#cc0000>X</font>)
<BR><FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>H</font>))
<BR><FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>R_IN_A</font>(<FONT COLOR=#cc0000>H</font>)
<BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>U4_A</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>T</font>))
<BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#cc0000>T</font>)
<BR><FONT COLOR=#0000cc>U1_A</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>U2_A</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>U1_A</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>Q_IN_G</font>(<FONT COLOR=#cc0000>X</font>)</BLOCKQUOTE><BR>The TRS R consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>p_in_a</font>(<FONT COLOR=#cc0000>X</font>) &#8594; <FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#0000cc>[]</font>) &#8594; <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#0000cc>[]</font>)
<BR><FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>H</font>))
<BR><FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#0000cc>1</font>) &#8594; <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)
<BR><FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>T</font>))
<BR><FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>))
<BR><FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>A</font>, <FONT COLOR=#0000cc>[]</font>)) &#8594; <FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>A</font>, <FONT COLOR=#0000cc>[]</font>))
<BR><FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>X</font>)</BLOCKQUOTE><BR>The argument filtering Pi contains the following mapping:<BR><FONT COLOR=#0000cc>p_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>p_in_a</font><BR>
<FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>l_in_a</font><BR>
<FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>r_in_a</font><BR>
<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>[]</font>&#160; = &#160;<FONT COLOR=#0000cc>[]</font><BR>
<FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>R_IN_A</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>R_IN_A</font><BR>
<FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>U1_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U1_A</font>(<FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>U4_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U4_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>U2_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U2_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>L_IN_A</font><BR>
<FONT COLOR=#0000cc>P_IN_A</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>P_IN_A</font><BR>
<FONT COLOR=#0000cc>Q_IN_G</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>Q_IN_G</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<BR>We have to consider all (P,R,Pi)-chains<BR><BR><pre>&#8627 Prolog</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>    &#8627 PiTRS</pre><pre>      &#8627 DependencyPairsProof</pre><pre>        &#8627 <B>PiDP</B></pre><pre>          &#8627 DependencyGraphProof</pre><BR>Pi DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>P_IN_A</font>(<FONT COLOR=#cc0000>X</font>) &#8594; <FONT COLOR=#0000cc>U1_A</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>P_IN_A</font>(<FONT COLOR=#cc0000>X</font>) &#8594; <FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#cc0000>X</font>)
<BR><FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>H</font>))
<BR><FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>R_IN_A</font>(<FONT COLOR=#cc0000>H</font>)
<BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>U4_A</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>T</font>))
<BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#cc0000>T</font>)
<BR><FONT COLOR=#0000cc>U1_A</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>U2_A</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>U1_A</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>Q_IN_G</font>(<FONT COLOR=#cc0000>X</font>)</BLOCKQUOTE><BR>The TRS R consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>p_in_a</font>(<FONT COLOR=#cc0000>X</font>) &#8594; <FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#0000cc>[]</font>) &#8594; <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#0000cc>[]</font>)
<BR><FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>H</font>))
<BR><FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#0000cc>1</font>) &#8594; <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)
<BR><FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>T</font>))
<BR><FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>))
<BR><FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>A</font>, <FONT COLOR=#0000cc>[]</font>)) &#8594; <FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>A</font>, <FONT COLOR=#0000cc>[]</font>))
<BR><FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>X</font>)</BLOCKQUOTE><BR>The argument filtering Pi contains the following mapping:<BR><FONT COLOR=#0000cc>p_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>p_in_a</font><BR>
<FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>l_in_a</font><BR>
<FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>r_in_a</font><BR>
<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>[]</font>&#160; = &#160;<FONT COLOR=#0000cc>[]</font><BR>
<FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>R_IN_A</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>R_IN_A</font><BR>
<FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>U1_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U1_A</font>(<FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>U4_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U4_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>U2_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U2_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>L_IN_A</font><BR>
<FONT COLOR=#0000cc>P_IN_A</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>P_IN_A</font><BR>
<FONT COLOR=#0000cc>Q_IN_G</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>Q_IN_G</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<BR>We have to consider all (P,R,Pi)-chains<BR>The approximation of the Dependency Graph [30] contains 1 SCC with 6 less nodes.<BR><BR><pre>&#8627 Prolog</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>    &#8627 PiTRS</pre><pre>      &#8627 DependencyPairsProof</pre><pre>        &#8627 PiDP</pre><pre>          &#8627 DependencyGraphProof</pre><pre>            &#8627 <B>PiDP</B></pre><pre>              &#8627 UsableRulesProof</pre><BR>Pi DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#cc0000>T</font>)
<BR><FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>H</font>))</BLOCKQUOTE><BR>The TRS R consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>p_in_a</font>(<FONT COLOR=#cc0000>X</font>) &#8594; <FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#0000cc>[]</font>) &#8594; <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#0000cc>[]</font>)
<BR><FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>H</font>))
<BR><FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#0000cc>1</font>) &#8594; <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)
<BR><FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>T</font>))
<BR><FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>))
<BR><FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>X</font>))
<BR><FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>A</font>, <FONT COLOR=#0000cc>[]</font>)) &#8594; <FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>A</font>, <FONT COLOR=#0000cc>[]</font>))
<BR><FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>X</font>, <FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>X</font>)) &#8594; <FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>X</font>)</BLOCKQUOTE><BR>The argument filtering Pi contains the following mapping:<BR><FONT COLOR=#0000cc>p_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>p_in_a</font><BR>
<FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U1_a</font>(<FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>l_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>l_in_a</font><BR>
<FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>l_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U3_a</font>(<FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>r_in_a</font><BR>
<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U4_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>U2_a</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>q_in_g</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>[]</font>&#160; = &#160;<FONT COLOR=#0000cc>[]</font><BR>
<FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>q_out_g</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>p_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>L_IN_A</font><BR>
<BR>We have to consider all (P,R,Pi)-chains<BR>For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.<BR><BR><pre>&#8627 Prolog</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>    &#8627 PiTRS</pre><pre>      &#8627 DependencyPairsProof</pre><pre>        &#8627 PiDP</pre><pre>          &#8627 DependencyGraphProof</pre><pre>            &#8627 PiDP</pre><pre>              &#8627 UsableRulesProof</pre><pre>                &#8627 <B>PiDP</B></pre><pre>                  &#8627 PiDPToQDPProof</pre><BR>Pi DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#cc0000>T</font>)
<BR><FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>)) &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>H</font>, <FONT COLOR=#cc0000>T</font>, <FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>H</font>))</BLOCKQUOTE><BR>The TRS R consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#0000cc>1</font>) &#8594; <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)</BLOCKQUOTE><BR>The argument filtering Pi contains the following mapping:<BR><FONT COLOR=#0000cc>r_in_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>r_in_a</font><BR>
<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>x1</font>)<BR>
<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)&#160; = &#160;<FONT COLOR=#0000cc>.</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>)<BR>
<FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>x1</font>, <FONT COLOR=#cc0000>x2</font>, <FONT COLOR=#cc0000>x3</font>)&#160; = &#160;<FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#cc0000>x3</font>)<BR>
<FONT COLOR=#0000cc>L_IN_A</font>(<FONT COLOR=#cc0000>x1</font>)&#160; = &#160;<FONT COLOR=#0000cc>L_IN_A</font><BR>
<BR>We have to consider all (P,R,Pi)-chains<BR>Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.<BR><BR><pre>&#8627 Prolog</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>    &#8627 PiTRS</pre><pre>      &#8627 DependencyPairsProof</pre><pre>        &#8627 PiDP</pre><pre>          &#8627 DependencyGraphProof</pre><pre>            &#8627 PiDP</pre><pre>              &#8627 UsableRulesProof</pre><pre>                &#8627 PiDP</pre><pre>                  &#8627 PiDPToQDPProof</pre><pre>                    &#8627 <B>QDP</B></pre><pre>                      &#8627 Rewriting</pre><BR>Q DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>L_IN_A</font> &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_in_a</font>)
<BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font></BLOCKQUOTE><BR>The TRS R consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>r_in_a</font> &#8594; <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)</BLOCKQUOTE><BR>The set Q consists of the following terms:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>r_in_a</font></BLOCKQUOTE><BR>We have to consider all (P,Q,R)-chains.<BR>By rewriting [15] the rule <FONT COLOR=#0000cc>L_IN_A</font> &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_in_a</font>) at position [0] we obtained the following new rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>L_IN_A</font> &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>))</BLOCKQUOTE><BR><BR><BR><pre>&#8627 Prolog</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>    &#8627 PiTRS</pre><pre>      &#8627 DependencyPairsProof</pre><pre>        &#8627 PiDP</pre><pre>          &#8627 DependencyGraphProof</pre><pre>            &#8627 PiDP</pre><pre>              &#8627 UsableRulesProof</pre><pre>                &#8627 PiDP</pre><pre>                  &#8627 PiDPToQDPProof</pre><pre>                    &#8627 QDP</pre><pre>                      &#8627 Rewriting</pre><pre>                        &#8627 <B>QDP</B></pre><pre>                          &#8627 UsableRulesProof</pre><BR>Q DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>L_IN_A</font> &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>))
<BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font></BLOCKQUOTE><BR>The TRS R consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>r_in_a</font> &#8594; <FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)</BLOCKQUOTE><BR>The set Q consists of the following terms:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>r_in_a</font></BLOCKQUOTE><BR>We have to consider all (P,Q,R)-chains.<BR>As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.<BR><BR><pre>&#8627 Prolog</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>    &#8627 PiTRS</pre><pre>      &#8627 DependencyPairsProof</pre><pre>        &#8627 PiDP</pre><pre>          &#8627 DependencyGraphProof</pre><pre>            &#8627 PiDP</pre><pre>              &#8627 UsableRulesProof</pre><pre>                &#8627 PiDP</pre><pre>                  &#8627 PiDPToQDPProof</pre><pre>                    &#8627 QDP</pre><pre>                      &#8627 Rewriting</pre><pre>                        &#8627 QDP</pre><pre>                          &#8627 UsableRulesProof</pre><pre>                            &#8627 <B>QDP</B></pre><pre>                              &#8627 QReductionProof</pre><BR>Q DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>L_IN_A</font> &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>))
<BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font></BLOCKQUOTE><BR>R is empty.<BR>The set Q consists of the following terms:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>r_in_a</font></BLOCKQUOTE><BR>We have to consider all (P,Q,R)-chains.<BR>We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>r_in_a</font></BLOCKQUOTE><BR><BR><BR><pre>&#8627 Prolog</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>    &#8627 PiTRS</pre><pre>      &#8627 DependencyPairsProof</pre><pre>        &#8627 PiDP</pre><pre>          &#8627 DependencyGraphProof</pre><pre>            &#8627 PiDP</pre><pre>              &#8627 UsableRulesProof</pre><pre>                &#8627 PiDP</pre><pre>                  &#8627 PiDPToQDPProof</pre><pre>                    &#8627 QDP</pre><pre>                      &#8627 Rewriting</pre><pre>                        &#8627 QDP</pre><pre>                          &#8627 UsableRulesProof</pre><pre>                            &#8627 QDP</pre><pre>                              &#8627 QReductionProof</pre><pre>                                &#8627 <B>QDP</B></pre><pre>                                  &#8627 Instantiation</pre><BR>Q DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>L_IN_A</font> &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>))
<BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font></BLOCKQUOTE><BR>R is empty.<BR>Q is empty.<BR>We have to consider all (P,Q,R)-chains.<BR>By instantiating [15] the rule <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#cc0000>H</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font> we obtained the following new rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font></BLOCKQUOTE><BR><BR><BR><pre>&#8627 Prolog</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>  &#8627 PrologToPiTRSProof</pre><pre>    &#8627 PiTRS</pre><pre>      &#8627 DependencyPairsProof</pre><pre>        &#8627 PiDP</pre><pre>          &#8627 DependencyGraphProof</pre><pre>            &#8627 PiDP</pre><pre>              &#8627 UsableRulesProof</pre><pre>                &#8627 PiDP</pre><pre>                  &#8627 PiDPToQDPProof</pre><pre>                    &#8627 QDP</pre><pre>                      &#8627 Rewriting</pre><pre>                        &#8627 QDP</pre><pre>                          &#8627 UsableRulesProof</pre><pre>                            &#8627 QDP</pre><pre>                              &#8627 QReductionProof</pre><pre>                                &#8627 QDP</pre><pre>                                  &#8627 Instantiation</pre><pre>                                    &#8627 <B>QDP</B></pre><pre>                                      &#8627 NonTerminationProof</pre><BR>Q DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>L_IN_A</font> &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>))
<BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font></BLOCKQUOTE><BR>R is empty.<BR>Q is empty.<BR>We have to consider all (P,Q,R)-chains.<BR>We used the non-termination processor [17] to show that the DP problem is infinite.<BR>Found a loop by narrowing to the left:<BR><BR>The TRS P consists of the following rules:<BLOCKQUOTE><BR><FONT COLOR=#0000cc>L_IN_A</font> &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>))
<BR><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font></BLOCKQUOTE><BR>The TRS R consists of the following rules:none<BR><BR><BR>s = <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)) evaluates to  t =<FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>))<BR><BR>Thus s starts an infinite chain as s semiunifies with t with the following substitutions:<BR><UL><LI> Semiunifier: [ ]</LI>
<LI> Matcher: [ ]</LI></UL><BR><BR><hr><BR><B>Rewriting sequence</B><BR><BR><B><FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>))</B> &#8594; <FONT COLOR=#0000cc>L_IN_A</font><BR>with rule <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>)) &#8594; <FONT COLOR=#0000cc>L_IN_A</font> at position [] and matcher [ ]<BR><BR><B><FONT COLOR=#0000cc>L_IN_A</font></B> &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>))<BR>with rule <FONT COLOR=#0000cc>L_IN_A</font> &#8594; <FONT COLOR=#0000cc>U3_A</font>(<FONT COLOR=#0000cc>r_out_a</font>(<FONT COLOR=#0000cc>1</font>))<BR><BR>Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence<BR><BR><BR>All these steps are and every following step will be a correct step w.r.t to Q.<BR><BR><BR><BR><BR><BR></body>

