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<BR><B>Left Termination</B> of the query pattern

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>a&#160;:-&#160;b.<BR>a&#160;:-&#160;e.<BR>b&#160;:-&#160;c.<BR>c&#160;:-&#160;d.<BR>d&#160;:-&#160;b.<BR>e&#160;:-&#160;f.<BR>f&#160;:-&#160;g.<BR>g&#160;:-&#160;e.<BR><BR>Queries:<BR><BR>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>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>a_in_</font> &#8594; <FONT COLOR=#0000cc>U1_</font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>b_in_</font> &#8594; <FONT COLOR=#0000cc>U3_</font>(<FONT COLOR=#0000cc>c_in_</font>)
<BR><FONT COLOR=#0000cc>c_in_</font> &#8594; <FONT COLOR=#0000cc>U4_</font>(<FONT COLOR=#0000cc>d_in_</font>)
<BR><FONT COLOR=#0000cc>d_in_</font> &#8594; <FONT COLOR=#0000cc>U5_</font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>U5_</font>(<FONT COLOR=#0000cc>b_out_</font>) &#8594; <FONT COLOR=#0000cc>d_out_</font>
<BR><FONT COLOR=#0000cc>U4_</font>(<FONT COLOR=#0000cc>d_out_</font>) &#8594; <FONT COLOR=#0000cc>c_out_</font>
<BR><FONT COLOR=#0000cc>U3_</font>(<FONT COLOR=#0000cc>c_out_</font>) &#8594; <FONT COLOR=#0000cc>b_out_</font>
<BR><FONT COLOR=#0000cc>U1_</font>(<FONT COLOR=#0000cc>b_out_</font>) &#8594; <FONT COLOR=#0000cc>a_out_</font>
<BR><FONT COLOR=#0000cc>a_in_</font> &#8594; <FONT COLOR=#0000cc>U2_</font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>e_in_</font> &#8594; <FONT COLOR=#0000cc>U6_</font>(<FONT COLOR=#0000cc>f_in_</font>)
<BR><FONT COLOR=#0000cc>f_in_</font> &#8594; <FONT COLOR=#0000cc>U7_</font>(<FONT COLOR=#0000cc>g_in_</font>)
<BR><FONT COLOR=#0000cc>g_in_</font> &#8594; <FONT COLOR=#0000cc>U8_</font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>U8_</font>(<FONT COLOR=#0000cc>e_out_</font>) &#8594; <FONT COLOR=#0000cc>g_out_</font>
<BR><FONT COLOR=#0000cc>U7_</font>(<FONT COLOR=#0000cc>g_out_</font>) &#8594; <FONT COLOR=#0000cc>f_out_</font>
<BR><FONT COLOR=#0000cc>U6_</font>(<FONT COLOR=#0000cc>f_out_</font>) &#8594; <FONT COLOR=#0000cc>e_out_</font>
<BR><FONT COLOR=#0000cc>U2_</font>(<FONT COLOR=#0000cc>e_out_</font>) &#8594; <FONT COLOR=#0000cc>a_out_</font></BLOCKQUOTE><BR>Pi is empty.<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>a_in_</font> &#8594; <FONT COLOR=#0000cc>U1_</font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>b_in_</font> &#8594; <FONT COLOR=#0000cc>U3_</font>(<FONT COLOR=#0000cc>c_in_</font>)
<BR><FONT COLOR=#0000cc>c_in_</font> &#8594; <FONT COLOR=#0000cc>U4_</font>(<FONT COLOR=#0000cc>d_in_</font>)
<BR><FONT COLOR=#0000cc>d_in_</font> &#8594; <FONT COLOR=#0000cc>U5_</font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>U5_</font>(<FONT COLOR=#0000cc>b_out_</font>) &#8594; <FONT COLOR=#0000cc>d_out_</font>
<BR><FONT COLOR=#0000cc>U4_</font>(<FONT COLOR=#0000cc>d_out_</font>) &#8594; <FONT COLOR=#0000cc>c_out_</font>
<BR><FONT COLOR=#0000cc>U3_</font>(<FONT COLOR=#0000cc>c_out_</font>) &#8594; <FONT COLOR=#0000cc>b_out_</font>
<BR><FONT COLOR=#0000cc>U1_</font>(<FONT COLOR=#0000cc>b_out_</font>) &#8594; <FONT COLOR=#0000cc>a_out_</font>
<BR><FONT COLOR=#0000cc>a_in_</font> &#8594; <FONT COLOR=#0000cc>U2_</font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>e_in_</font> &#8594; <FONT COLOR=#0000cc>U6_</font>(<FONT COLOR=#0000cc>f_in_</font>)
<BR><FONT COLOR=#0000cc>f_in_</font> &#8594; <FONT COLOR=#0000cc>U7_</font>(<FONT COLOR=#0000cc>g_in_</font>)
<BR><FONT COLOR=#0000cc>g_in_</font> &#8594; <FONT COLOR=#0000cc>U8_</font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>U8_</font>(<FONT COLOR=#0000cc>e_out_</font>) &#8594; <FONT COLOR=#0000cc>g_out_</font>
<BR><FONT COLOR=#0000cc>U7_</font>(<FONT COLOR=#0000cc>g_out_</font>) &#8594; <FONT COLOR=#0000cc>f_out_</font>
<BR><FONT COLOR=#0000cc>U6_</font>(<FONT COLOR=#0000cc>f_out_</font>) &#8594; <FONT COLOR=#0000cc>e_out_</font>
<BR><FONT COLOR=#0000cc>U2_</font>(<FONT COLOR=#0000cc>e_out_</font>) &#8594; <FONT COLOR=#0000cc>a_out_</font></BLOCKQUOTE><BR>Pi is empty.<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>A_IN_</font> &#8594; <FONT COLOR=#0000cc>U1_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>A_IN_</font> &#8594; <FONT COLOR=#0000cc>B_IN_</font>
<BR><FONT COLOR=#0000cc>B_IN_</font> &#8594; <FONT COLOR=#0000cc>U3_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>c_in_</font>)
<BR><FONT COLOR=#0000cc>B_IN_</font> &#8594; <FONT COLOR=#0000cc>C_IN_</font>
<BR><FONT COLOR=#0000cc>C_IN_</font> &#8594; <FONT COLOR=#0000cc>U4_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>d_in_</font>)
<BR><FONT COLOR=#0000cc>C_IN_</font> &#8594; <FONT COLOR=#0000cc>D_IN_</font>
<BR><FONT COLOR=#0000cc>D_IN_</font> &#8594; <FONT COLOR=#0000cc>U5_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>D_IN_</font> &#8594; <FONT COLOR=#0000cc>B_IN_</font>
<BR><FONT COLOR=#0000cc>A_IN_</font> &#8594; <FONT COLOR=#0000cc>U2_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>A_IN_</font> &#8594; <FONT COLOR=#0000cc>E_IN_</font>
<BR><FONT COLOR=#0000cc>E_IN_</font> &#8594; <FONT COLOR=#0000cc>U6_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>f_in_</font>)
<BR><FONT COLOR=#0000cc>E_IN_</font> &#8594; <FONT COLOR=#0000cc>F_IN_</font>
<BR><FONT COLOR=#0000cc>F_IN_</font> &#8594; <FONT COLOR=#0000cc>U7_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>g_in_</font>)
<BR><FONT COLOR=#0000cc>F_IN_</font> &#8594; <FONT COLOR=#0000cc>G_IN_</font>
<BR><FONT COLOR=#0000cc>G_IN_</font> &#8594; <FONT COLOR=#0000cc>U8_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>G_IN_</font> &#8594; <FONT COLOR=#0000cc>E_IN_</font></BLOCKQUOTE><BR>The TRS R consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>a_in_</font> &#8594; <FONT COLOR=#0000cc>U1_</font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>b_in_</font> &#8594; <FONT COLOR=#0000cc>U3_</font>(<FONT COLOR=#0000cc>c_in_</font>)
<BR><FONT COLOR=#0000cc>c_in_</font> &#8594; <FONT COLOR=#0000cc>U4_</font>(<FONT COLOR=#0000cc>d_in_</font>)
<BR><FONT COLOR=#0000cc>d_in_</font> &#8594; <FONT COLOR=#0000cc>U5_</font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>U5_</font>(<FONT COLOR=#0000cc>b_out_</font>) &#8594; <FONT COLOR=#0000cc>d_out_</font>
<BR><FONT COLOR=#0000cc>U4_</font>(<FONT COLOR=#0000cc>d_out_</font>) &#8594; <FONT COLOR=#0000cc>c_out_</font>
<BR><FONT COLOR=#0000cc>U3_</font>(<FONT COLOR=#0000cc>c_out_</font>) &#8594; <FONT COLOR=#0000cc>b_out_</font>
<BR><FONT COLOR=#0000cc>U1_</font>(<FONT COLOR=#0000cc>b_out_</font>) &#8594; <FONT COLOR=#0000cc>a_out_</font>
<BR><FONT COLOR=#0000cc>a_in_</font> &#8594; <FONT COLOR=#0000cc>U2_</font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>e_in_</font> &#8594; <FONT COLOR=#0000cc>U6_</font>(<FONT COLOR=#0000cc>f_in_</font>)
<BR><FONT COLOR=#0000cc>f_in_</font> &#8594; <FONT COLOR=#0000cc>U7_</font>(<FONT COLOR=#0000cc>g_in_</font>)
<BR><FONT COLOR=#0000cc>g_in_</font> &#8594; <FONT COLOR=#0000cc>U8_</font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>U8_</font>(<FONT COLOR=#0000cc>e_out_</font>) &#8594; <FONT COLOR=#0000cc>g_out_</font>
<BR><FONT COLOR=#0000cc>U7_</font>(<FONT COLOR=#0000cc>g_out_</font>) &#8594; <FONT COLOR=#0000cc>f_out_</font>
<BR><FONT COLOR=#0000cc>U6_</font>(<FONT COLOR=#0000cc>f_out_</font>) &#8594; <FONT COLOR=#0000cc>e_out_</font>
<BR><FONT COLOR=#0000cc>U2_</font>(<FONT COLOR=#0000cc>e_out_</font>) &#8594; <FONT COLOR=#0000cc>a_out_</font></BLOCKQUOTE><BR>Pi is empty.<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>A_IN_</font> &#8594; <FONT COLOR=#0000cc>U1_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>A_IN_</font> &#8594; <FONT COLOR=#0000cc>B_IN_</font>
<BR><FONT COLOR=#0000cc>B_IN_</font> &#8594; <FONT COLOR=#0000cc>U3_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>c_in_</font>)
<BR><FONT COLOR=#0000cc>B_IN_</font> &#8594; <FONT COLOR=#0000cc>C_IN_</font>
<BR><FONT COLOR=#0000cc>C_IN_</font> &#8594; <FONT COLOR=#0000cc>U4_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>d_in_</font>)
<BR><FONT COLOR=#0000cc>C_IN_</font> &#8594; <FONT COLOR=#0000cc>D_IN_</font>
<BR><FONT COLOR=#0000cc>D_IN_</font> &#8594; <FONT COLOR=#0000cc>U5_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>D_IN_</font> &#8594; <FONT COLOR=#0000cc>B_IN_</font>
<BR><FONT COLOR=#0000cc>A_IN_</font> &#8594; <FONT COLOR=#0000cc>U2_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>A_IN_</font> &#8594; <FONT COLOR=#0000cc>E_IN_</font>
<BR><FONT COLOR=#0000cc>E_IN_</font> &#8594; <FONT COLOR=#0000cc>U6_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>f_in_</font>)
<BR><FONT COLOR=#0000cc>E_IN_</font> &#8594; <FONT COLOR=#0000cc>F_IN_</font>
<BR><FONT COLOR=#0000cc>F_IN_</font> &#8594; <FONT COLOR=#0000cc>U7_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>g_in_</font>)
<BR><FONT COLOR=#0000cc>F_IN_</font> &#8594; <FONT COLOR=#0000cc>G_IN_</font>
<BR><FONT COLOR=#0000cc>G_IN_</font> &#8594; <FONT COLOR=#0000cc>U8_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>G_IN_</font> &#8594; <FONT COLOR=#0000cc>E_IN_</font></BLOCKQUOTE><BR>The TRS R consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>a_in_</font> &#8594; <FONT COLOR=#0000cc>U1_</font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>b_in_</font> &#8594; <FONT COLOR=#0000cc>U3_</font>(<FONT COLOR=#0000cc>c_in_</font>)
<BR><FONT COLOR=#0000cc>c_in_</font> &#8594; <FONT COLOR=#0000cc>U4_</font>(<FONT COLOR=#0000cc>d_in_</font>)
<BR><FONT COLOR=#0000cc>d_in_</font> &#8594; <FONT COLOR=#0000cc>U5_</font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>U5_</font>(<FONT COLOR=#0000cc>b_out_</font>) &#8594; <FONT COLOR=#0000cc>d_out_</font>
<BR><FONT COLOR=#0000cc>U4_</font>(<FONT COLOR=#0000cc>d_out_</font>) &#8594; <FONT COLOR=#0000cc>c_out_</font>
<BR><FONT COLOR=#0000cc>U3_</font>(<FONT COLOR=#0000cc>c_out_</font>) &#8594; <FONT COLOR=#0000cc>b_out_</font>
<BR><FONT COLOR=#0000cc>U1_</font>(<FONT COLOR=#0000cc>b_out_</font>) &#8594; <FONT COLOR=#0000cc>a_out_</font>
<BR><FONT COLOR=#0000cc>a_in_</font> &#8594; <FONT COLOR=#0000cc>U2_</font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>e_in_</font> &#8594; <FONT COLOR=#0000cc>U6_</font>(<FONT COLOR=#0000cc>f_in_</font>)
<BR><FONT COLOR=#0000cc>f_in_</font> &#8594; <FONT COLOR=#0000cc>U7_</font>(<FONT COLOR=#0000cc>g_in_</font>)
<BR><FONT COLOR=#0000cc>g_in_</font> &#8594; <FONT COLOR=#0000cc>U8_</font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>U8_</font>(<FONT COLOR=#0000cc>e_out_</font>) &#8594; <FONT COLOR=#0000cc>g_out_</font>
<BR><FONT COLOR=#0000cc>U7_</font>(<FONT COLOR=#0000cc>g_out_</font>) &#8594; <FONT COLOR=#0000cc>f_out_</font>
<BR><FONT COLOR=#0000cc>U6_</font>(<FONT COLOR=#0000cc>f_out_</font>) &#8594; <FONT COLOR=#0000cc>e_out_</font>
<BR><FONT COLOR=#0000cc>U2_</font>(<FONT COLOR=#0000cc>e_out_</font>) &#8594; <FONT COLOR=#0000cc>a_out_</font></BLOCKQUOTE><BR>Pi is empty.<BR>We have to consider all (P,R,Pi)-chains<BR>The approximation of the Dependency Graph [30] contains 2 SCCs with 10 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 AND</pre><pre>              &#8627 <B>PiDP</B></pre><pre>                &#8627 UsableRulesProof</pre><pre>              &#8627 PiDP</pre><pre>  &#8627 PrologToPiTRSProof</pre><BR>Pi DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>E_IN_</font> &#8594; <FONT COLOR=#0000cc>F_IN_</font>
<BR><FONT COLOR=#0000cc>F_IN_</font> &#8594; <FONT COLOR=#0000cc>G_IN_</font>
<BR><FONT COLOR=#0000cc>G_IN_</font> &#8594; <FONT COLOR=#0000cc>E_IN_</font></BLOCKQUOTE><BR>The TRS R consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>a_in_</font> &#8594; <FONT COLOR=#0000cc>U1_</font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>b_in_</font> &#8594; <FONT COLOR=#0000cc>U3_</font>(<FONT COLOR=#0000cc>c_in_</font>)
<BR><FONT COLOR=#0000cc>c_in_</font> &#8594; <FONT COLOR=#0000cc>U4_</font>(<FONT COLOR=#0000cc>d_in_</font>)
<BR><FONT COLOR=#0000cc>d_in_</font> &#8594; <FONT COLOR=#0000cc>U5_</font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>U5_</font>(<FONT COLOR=#0000cc>b_out_</font>) &#8594; <FONT COLOR=#0000cc>d_out_</font>
<BR><FONT COLOR=#0000cc>U4_</font>(<FONT COLOR=#0000cc>d_out_</font>) &#8594; <FONT COLOR=#0000cc>c_out_</font>
<BR><FONT COLOR=#0000cc>U3_</font>(<FONT COLOR=#0000cc>c_out_</font>) &#8594; <FONT COLOR=#0000cc>b_out_</font>
<BR><FONT COLOR=#0000cc>U1_</font>(<FONT COLOR=#0000cc>b_out_</font>) &#8594; <FONT COLOR=#0000cc>a_out_</font>
<BR><FONT COLOR=#0000cc>a_in_</font> &#8594; <FONT COLOR=#0000cc>U2_</font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>e_in_</font> &#8594; <FONT COLOR=#0000cc>U6_</font>(<FONT COLOR=#0000cc>f_in_</font>)
<BR><FONT COLOR=#0000cc>f_in_</font> &#8594; <FONT COLOR=#0000cc>U7_</font>(<FONT COLOR=#0000cc>g_in_</font>)
<BR><FONT COLOR=#0000cc>g_in_</font> &#8594; <FONT COLOR=#0000cc>U8_</font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>U8_</font>(<FONT COLOR=#0000cc>e_out_</font>) &#8594; <FONT COLOR=#0000cc>g_out_</font>
<BR><FONT COLOR=#0000cc>U7_</font>(<FONT COLOR=#0000cc>g_out_</font>) &#8594; <FONT COLOR=#0000cc>f_out_</font>
<BR><FONT COLOR=#0000cc>U6_</font>(<FONT COLOR=#0000cc>f_out_</font>) &#8594; <FONT COLOR=#0000cc>e_out_</font>
<BR><FONT COLOR=#0000cc>U2_</font>(<FONT COLOR=#0000cc>e_out_</font>) &#8594; <FONT COLOR=#0000cc>a_out_</font></BLOCKQUOTE><BR>Pi is empty.<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 AND</pre><pre>              &#8627 PiDP</pre><pre>                &#8627 UsableRulesProof</pre><pre>                  &#8627 <B>PiDP</B></pre><pre>                    &#8627 PiDPToQDPProof</pre><pre>              &#8627 PiDP</pre><pre>  &#8627 PrologToPiTRSProof</pre><BR>Pi DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>E_IN_</font> &#8594; <FONT COLOR=#0000cc>F_IN_</font>
<BR><FONT COLOR=#0000cc>F_IN_</font> &#8594; <FONT COLOR=#0000cc>G_IN_</font>
<BR><FONT COLOR=#0000cc>G_IN_</font> &#8594; <FONT COLOR=#0000cc>E_IN_</font></BLOCKQUOTE><BR>R is empty.<BR>Pi is empty.<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 AND</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 NonTerminationProof</pre><pre>              &#8627 PiDP</pre><pre>  &#8627 PrologToPiTRSProof</pre><BR>Q DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>E_IN_</font> &#8594; <FONT COLOR=#0000cc>F_IN_</font>
<BR><FONT COLOR=#0000cc>F_IN_</font> &#8594; <FONT COLOR=#0000cc>G_IN_</font>
<BR><FONT COLOR=#0000cc>G_IN_</font> &#8594; <FONT COLOR=#0000cc>E_IN_</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>E_IN_</font> &#8594; <FONT COLOR=#0000cc>F_IN_</font>
<BR><FONT COLOR=#0000cc>F_IN_</font> &#8594; <FONT COLOR=#0000cc>G_IN_</font>
<BR><FONT COLOR=#0000cc>G_IN_</font> &#8594; <FONT COLOR=#0000cc>E_IN_</font></BLOCKQUOTE><BR>The TRS R consists of the following rules:none<BR><BR><BR>s = <FONT COLOR=#0000cc>F_IN_</font> evaluates to  t =<FONT COLOR=#0000cc>F_IN_</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>F_IN_</font></B> &#8594; <FONT COLOR=#0000cc>G_IN_</font><BR>with rule <FONT COLOR=#0000cc>F_IN_</font> &#8594; <FONT COLOR=#0000cc>G_IN_</font> at position [] and matcher [ ]<BR><BR><B><FONT COLOR=#0000cc>G_IN_</font></B> &#8594; <FONT COLOR=#0000cc>E_IN_</font><BR>with rule <FONT COLOR=#0000cc>G_IN_</font> &#8594; <FONT COLOR=#0000cc>E_IN_</font> at position [] and matcher [ ]<BR><BR><B><FONT COLOR=#0000cc>E_IN_</font></B> &#8594; <FONT COLOR=#0000cc>F_IN_</font><BR>with rule <FONT COLOR=#0000cc>E_IN_</font> &#8594; <FONT COLOR=#0000cc>F_IN_</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><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 AND</pre><pre>              &#8627 PiDP</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>C_IN_</font> &#8594; <FONT COLOR=#0000cc>D_IN_</font>
<BR><FONT COLOR=#0000cc>B_IN_</font> &#8594; <FONT COLOR=#0000cc>C_IN_</font>
<BR><FONT COLOR=#0000cc>D_IN_</font> &#8594; <FONT COLOR=#0000cc>B_IN_</font></BLOCKQUOTE><BR>The TRS R consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>a_in_</font> &#8594; <FONT COLOR=#0000cc>U1_</font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>b_in_</font> &#8594; <FONT COLOR=#0000cc>U3_</font>(<FONT COLOR=#0000cc>c_in_</font>)
<BR><FONT COLOR=#0000cc>c_in_</font> &#8594; <FONT COLOR=#0000cc>U4_</font>(<FONT COLOR=#0000cc>d_in_</font>)
<BR><FONT COLOR=#0000cc>d_in_</font> &#8594; <FONT COLOR=#0000cc>U5_</font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>U5_</font>(<FONT COLOR=#0000cc>b_out_</font>) &#8594; <FONT COLOR=#0000cc>d_out_</font>
<BR><FONT COLOR=#0000cc>U4_</font>(<FONT COLOR=#0000cc>d_out_</font>) &#8594; <FONT COLOR=#0000cc>c_out_</font>
<BR><FONT COLOR=#0000cc>U3_</font>(<FONT COLOR=#0000cc>c_out_</font>) &#8594; <FONT COLOR=#0000cc>b_out_</font>
<BR><FONT COLOR=#0000cc>U1_</font>(<FONT COLOR=#0000cc>b_out_</font>) &#8594; <FONT COLOR=#0000cc>a_out_</font>
<BR><FONT COLOR=#0000cc>a_in_</font> &#8594; <FONT COLOR=#0000cc>U2_</font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>e_in_</font> &#8594; <FONT COLOR=#0000cc>U6_</font>(<FONT COLOR=#0000cc>f_in_</font>)
<BR><FONT COLOR=#0000cc>f_in_</font> &#8594; <FONT COLOR=#0000cc>U7_</font>(<FONT COLOR=#0000cc>g_in_</font>)
<BR><FONT COLOR=#0000cc>g_in_</font> &#8594; <FONT COLOR=#0000cc>U8_</font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>U8_</font>(<FONT COLOR=#0000cc>e_out_</font>) &#8594; <FONT COLOR=#0000cc>g_out_</font>
<BR><FONT COLOR=#0000cc>U7_</font>(<FONT COLOR=#0000cc>g_out_</font>) &#8594; <FONT COLOR=#0000cc>f_out_</font>
<BR><FONT COLOR=#0000cc>U6_</font>(<FONT COLOR=#0000cc>f_out_</font>) &#8594; <FONT COLOR=#0000cc>e_out_</font>
<BR><FONT COLOR=#0000cc>U2_</font>(<FONT COLOR=#0000cc>e_out_</font>) &#8594; <FONT COLOR=#0000cc>a_out_</font></BLOCKQUOTE><BR>Pi is empty.<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 AND</pre><pre>              &#8627 PiDP</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>C_IN_</font> &#8594; <FONT COLOR=#0000cc>D_IN_</font>
<BR><FONT COLOR=#0000cc>B_IN_</font> &#8594; <FONT COLOR=#0000cc>C_IN_</font>
<BR><FONT COLOR=#0000cc>D_IN_</font> &#8594; <FONT COLOR=#0000cc>B_IN_</font></BLOCKQUOTE><BR>R is empty.<BR>Pi is empty.<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 AND</pre><pre>              &#8627 PiDP</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 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>C_IN_</font> &#8594; <FONT COLOR=#0000cc>D_IN_</font>
<BR><FONT COLOR=#0000cc>B_IN_</font> &#8594; <FONT COLOR=#0000cc>C_IN_</font>
<BR><FONT COLOR=#0000cc>D_IN_</font> &#8594; <FONT COLOR=#0000cc>B_IN_</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>C_IN_</font> &#8594; <FONT COLOR=#0000cc>D_IN_</font>
<BR><FONT COLOR=#0000cc>B_IN_</font> &#8594; <FONT COLOR=#0000cc>C_IN_</font>
<BR><FONT COLOR=#0000cc>D_IN_</font> &#8594; <FONT COLOR=#0000cc>B_IN_</font></BLOCKQUOTE><BR>The TRS R consists of the following rules:none<BR><BR><BR>s = <FONT COLOR=#0000cc>D_IN_</font> evaluates to  t =<FONT COLOR=#0000cc>D_IN_</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>D_IN_</font></B> &#8594; <FONT COLOR=#0000cc>B_IN_</font><BR>with rule <FONT COLOR=#0000cc>D_IN_</font> &#8594; <FONT COLOR=#0000cc>B_IN_</font> at position [] and matcher [ ]<BR><BR><B><FONT COLOR=#0000cc>B_IN_</font></B> &#8594; <FONT COLOR=#0000cc>C_IN_</font><BR>with rule <FONT COLOR=#0000cc>B_IN_</font> &#8594; <FONT COLOR=#0000cc>C_IN_</font> at position [] and matcher [ ]<BR><BR><B><FONT COLOR=#0000cc>C_IN_</font></B> &#8594; <FONT COLOR=#0000cc>D_IN_</font><BR>with rule <FONT COLOR=#0000cc>C_IN_</font> &#8594; <FONT COLOR=#0000cc>D_IN_</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>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>a_in_</font> &#8594; <FONT COLOR=#0000cc>U1_</font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>b_in_</font> &#8594; <FONT COLOR=#0000cc>U3_</font>(<FONT COLOR=#0000cc>c_in_</font>)
<BR><FONT COLOR=#0000cc>c_in_</font> &#8594; <FONT COLOR=#0000cc>U4_</font>(<FONT COLOR=#0000cc>d_in_</font>)
<BR><FONT COLOR=#0000cc>d_in_</font> &#8594; <FONT COLOR=#0000cc>U5_</font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>U5_</font>(<FONT COLOR=#0000cc>b_out_</font>) &#8594; <FONT COLOR=#0000cc>d_out_</font>
<BR><FONT COLOR=#0000cc>U4_</font>(<FONT COLOR=#0000cc>d_out_</font>) &#8594; <FONT COLOR=#0000cc>c_out_</font>
<BR><FONT COLOR=#0000cc>U3_</font>(<FONT COLOR=#0000cc>c_out_</font>) &#8594; <FONT COLOR=#0000cc>b_out_</font>
<BR><FONT COLOR=#0000cc>U1_</font>(<FONT COLOR=#0000cc>b_out_</font>) &#8594; <FONT COLOR=#0000cc>a_out_</font>
<BR><FONT COLOR=#0000cc>a_in_</font> &#8594; <FONT COLOR=#0000cc>U2_</font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>e_in_</font> &#8594; <FONT COLOR=#0000cc>U6_</font>(<FONT COLOR=#0000cc>f_in_</font>)
<BR><FONT COLOR=#0000cc>f_in_</font> &#8594; <FONT COLOR=#0000cc>U7_</font>(<FONT COLOR=#0000cc>g_in_</font>)
<BR><FONT COLOR=#0000cc>g_in_</font> &#8594; <FONT COLOR=#0000cc>U8_</font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>U8_</font>(<FONT COLOR=#0000cc>e_out_</font>) &#8594; <FONT COLOR=#0000cc>g_out_</font>
<BR><FONT COLOR=#0000cc>U7_</font>(<FONT COLOR=#0000cc>g_out_</font>) &#8594; <FONT COLOR=#0000cc>f_out_</font>
<BR><FONT COLOR=#0000cc>U6_</font>(<FONT COLOR=#0000cc>f_out_</font>) &#8594; <FONT COLOR=#0000cc>e_out_</font>
<BR><FONT COLOR=#0000cc>U2_</font>(<FONT COLOR=#0000cc>e_out_</font>) &#8594; <FONT COLOR=#0000cc>a_out_</font></BLOCKQUOTE><BR>Pi is empty.<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>a_in_</font> &#8594; <FONT COLOR=#0000cc>U1_</font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>b_in_</font> &#8594; <FONT COLOR=#0000cc>U3_</font>(<FONT COLOR=#0000cc>c_in_</font>)
<BR><FONT COLOR=#0000cc>c_in_</font> &#8594; <FONT COLOR=#0000cc>U4_</font>(<FONT COLOR=#0000cc>d_in_</font>)
<BR><FONT COLOR=#0000cc>d_in_</font> &#8594; <FONT COLOR=#0000cc>U5_</font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>U5_</font>(<FONT COLOR=#0000cc>b_out_</font>) &#8594; <FONT COLOR=#0000cc>d_out_</font>
<BR><FONT COLOR=#0000cc>U4_</font>(<FONT COLOR=#0000cc>d_out_</font>) &#8594; <FONT COLOR=#0000cc>c_out_</font>
<BR><FONT COLOR=#0000cc>U3_</font>(<FONT COLOR=#0000cc>c_out_</font>) &#8594; <FONT COLOR=#0000cc>b_out_</font>
<BR><FONT COLOR=#0000cc>U1_</font>(<FONT COLOR=#0000cc>b_out_</font>) &#8594; <FONT COLOR=#0000cc>a_out_</font>
<BR><FONT COLOR=#0000cc>a_in_</font> &#8594; <FONT COLOR=#0000cc>U2_</font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>e_in_</font> &#8594; <FONT COLOR=#0000cc>U6_</font>(<FONT COLOR=#0000cc>f_in_</font>)
<BR><FONT COLOR=#0000cc>f_in_</font> &#8594; <FONT COLOR=#0000cc>U7_</font>(<FONT COLOR=#0000cc>g_in_</font>)
<BR><FONT COLOR=#0000cc>g_in_</font> &#8594; <FONT COLOR=#0000cc>U8_</font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>U8_</font>(<FONT COLOR=#0000cc>e_out_</font>) &#8594; <FONT COLOR=#0000cc>g_out_</font>
<BR><FONT COLOR=#0000cc>U7_</font>(<FONT COLOR=#0000cc>g_out_</font>) &#8594; <FONT COLOR=#0000cc>f_out_</font>
<BR><FONT COLOR=#0000cc>U6_</font>(<FONT COLOR=#0000cc>f_out_</font>) &#8594; <FONT COLOR=#0000cc>e_out_</font>
<BR><FONT COLOR=#0000cc>U2_</font>(<FONT COLOR=#0000cc>e_out_</font>) &#8594; <FONT COLOR=#0000cc>a_out_</font></BLOCKQUOTE><BR>Pi is empty.<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>A_IN_</font> &#8594; <FONT COLOR=#0000cc>U1_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>A_IN_</font> &#8594; <FONT COLOR=#0000cc>B_IN_</font>
<BR><FONT COLOR=#0000cc>B_IN_</font> &#8594; <FONT COLOR=#0000cc>U3_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>c_in_</font>)
<BR><FONT COLOR=#0000cc>B_IN_</font> &#8594; <FONT COLOR=#0000cc>C_IN_</font>
<BR><FONT COLOR=#0000cc>C_IN_</font> &#8594; <FONT COLOR=#0000cc>U4_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>d_in_</font>)
<BR><FONT COLOR=#0000cc>C_IN_</font> &#8594; <FONT COLOR=#0000cc>D_IN_</font>
<BR><FONT COLOR=#0000cc>D_IN_</font> &#8594; <FONT COLOR=#0000cc>U5_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>D_IN_</font> &#8594; <FONT COLOR=#0000cc>B_IN_</font>
<BR><FONT COLOR=#0000cc>A_IN_</font> &#8594; <FONT COLOR=#0000cc>U2_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>A_IN_</font> &#8594; <FONT COLOR=#0000cc>E_IN_</font>
<BR><FONT COLOR=#0000cc>E_IN_</font> &#8594; <FONT COLOR=#0000cc>U6_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>f_in_</font>)
<BR><FONT COLOR=#0000cc>E_IN_</font> &#8594; <FONT COLOR=#0000cc>F_IN_</font>
<BR><FONT COLOR=#0000cc>F_IN_</font> &#8594; <FONT COLOR=#0000cc>U7_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>g_in_</font>)
<BR><FONT COLOR=#0000cc>F_IN_</font> &#8594; <FONT COLOR=#0000cc>G_IN_</font>
<BR><FONT COLOR=#0000cc>G_IN_</font> &#8594; <FONT COLOR=#0000cc>U8_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>G_IN_</font> &#8594; <FONT COLOR=#0000cc>E_IN_</font></BLOCKQUOTE><BR>The TRS R consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>a_in_</font> &#8594; <FONT COLOR=#0000cc>U1_</font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>b_in_</font> &#8594; <FONT COLOR=#0000cc>U3_</font>(<FONT COLOR=#0000cc>c_in_</font>)
<BR><FONT COLOR=#0000cc>c_in_</font> &#8594; <FONT COLOR=#0000cc>U4_</font>(<FONT COLOR=#0000cc>d_in_</font>)
<BR><FONT COLOR=#0000cc>d_in_</font> &#8594; <FONT COLOR=#0000cc>U5_</font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>U5_</font>(<FONT COLOR=#0000cc>b_out_</font>) &#8594; <FONT COLOR=#0000cc>d_out_</font>
<BR><FONT COLOR=#0000cc>U4_</font>(<FONT COLOR=#0000cc>d_out_</font>) &#8594; <FONT COLOR=#0000cc>c_out_</font>
<BR><FONT COLOR=#0000cc>U3_</font>(<FONT COLOR=#0000cc>c_out_</font>) &#8594; <FONT COLOR=#0000cc>b_out_</font>
<BR><FONT COLOR=#0000cc>U1_</font>(<FONT COLOR=#0000cc>b_out_</font>) &#8594; <FONT COLOR=#0000cc>a_out_</font>
<BR><FONT COLOR=#0000cc>a_in_</font> &#8594; <FONT COLOR=#0000cc>U2_</font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>e_in_</font> &#8594; <FONT COLOR=#0000cc>U6_</font>(<FONT COLOR=#0000cc>f_in_</font>)
<BR><FONT COLOR=#0000cc>f_in_</font> &#8594; <FONT COLOR=#0000cc>U7_</font>(<FONT COLOR=#0000cc>g_in_</font>)
<BR><FONT COLOR=#0000cc>g_in_</font> &#8594; <FONT COLOR=#0000cc>U8_</font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>U8_</font>(<FONT COLOR=#0000cc>e_out_</font>) &#8594; <FONT COLOR=#0000cc>g_out_</font>
<BR><FONT COLOR=#0000cc>U7_</font>(<FONT COLOR=#0000cc>g_out_</font>) &#8594; <FONT COLOR=#0000cc>f_out_</font>
<BR><FONT COLOR=#0000cc>U6_</font>(<FONT COLOR=#0000cc>f_out_</font>) &#8594; <FONT COLOR=#0000cc>e_out_</font>
<BR><FONT COLOR=#0000cc>U2_</font>(<FONT COLOR=#0000cc>e_out_</font>) &#8594; <FONT COLOR=#0000cc>a_out_</font></BLOCKQUOTE><BR>Pi is empty.<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>A_IN_</font> &#8594; <FONT COLOR=#0000cc>U1_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>A_IN_</font> &#8594; <FONT COLOR=#0000cc>B_IN_</font>
<BR><FONT COLOR=#0000cc>B_IN_</font> &#8594; <FONT COLOR=#0000cc>U3_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>c_in_</font>)
<BR><FONT COLOR=#0000cc>B_IN_</font> &#8594; <FONT COLOR=#0000cc>C_IN_</font>
<BR><FONT COLOR=#0000cc>C_IN_</font> &#8594; <FONT COLOR=#0000cc>U4_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>d_in_</font>)
<BR><FONT COLOR=#0000cc>C_IN_</font> &#8594; <FONT COLOR=#0000cc>D_IN_</font>
<BR><FONT COLOR=#0000cc>D_IN_</font> &#8594; <FONT COLOR=#0000cc>U5_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>D_IN_</font> &#8594; <FONT COLOR=#0000cc>B_IN_</font>
<BR><FONT COLOR=#0000cc>A_IN_</font> &#8594; <FONT COLOR=#0000cc>U2_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>A_IN_</font> &#8594; <FONT COLOR=#0000cc>E_IN_</font>
<BR><FONT COLOR=#0000cc>E_IN_</font> &#8594; <FONT COLOR=#0000cc>U6_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>f_in_</font>)
<BR><FONT COLOR=#0000cc>E_IN_</font> &#8594; <FONT COLOR=#0000cc>F_IN_</font>
<BR><FONT COLOR=#0000cc>F_IN_</font> &#8594; <FONT COLOR=#0000cc>U7_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>g_in_</font>)
<BR><FONT COLOR=#0000cc>F_IN_</font> &#8594; <FONT COLOR=#0000cc>G_IN_</font>
<BR><FONT COLOR=#0000cc>G_IN_</font> &#8594; <FONT COLOR=#0000cc>U8_<SUP>1</SUP></font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>G_IN_</font> &#8594; <FONT COLOR=#0000cc>E_IN_</font></BLOCKQUOTE><BR>The TRS R consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>a_in_</font> &#8594; <FONT COLOR=#0000cc>U1_</font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>b_in_</font> &#8594; <FONT COLOR=#0000cc>U3_</font>(<FONT COLOR=#0000cc>c_in_</font>)
<BR><FONT COLOR=#0000cc>c_in_</font> &#8594; <FONT COLOR=#0000cc>U4_</font>(<FONT COLOR=#0000cc>d_in_</font>)
<BR><FONT COLOR=#0000cc>d_in_</font> &#8594; <FONT COLOR=#0000cc>U5_</font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>U5_</font>(<FONT COLOR=#0000cc>b_out_</font>) &#8594; <FONT COLOR=#0000cc>d_out_</font>
<BR><FONT COLOR=#0000cc>U4_</font>(<FONT COLOR=#0000cc>d_out_</font>) &#8594; <FONT COLOR=#0000cc>c_out_</font>
<BR><FONT COLOR=#0000cc>U3_</font>(<FONT COLOR=#0000cc>c_out_</font>) &#8594; <FONT COLOR=#0000cc>b_out_</font>
<BR><FONT COLOR=#0000cc>U1_</font>(<FONT COLOR=#0000cc>b_out_</font>) &#8594; <FONT COLOR=#0000cc>a_out_</font>
<BR><FONT COLOR=#0000cc>a_in_</font> &#8594; <FONT COLOR=#0000cc>U2_</font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>e_in_</font> &#8594; <FONT COLOR=#0000cc>U6_</font>(<FONT COLOR=#0000cc>f_in_</font>)
<BR><FONT COLOR=#0000cc>f_in_</font> &#8594; <FONT COLOR=#0000cc>U7_</font>(<FONT COLOR=#0000cc>g_in_</font>)
<BR><FONT COLOR=#0000cc>g_in_</font> &#8594; <FONT COLOR=#0000cc>U8_</font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>U8_</font>(<FONT COLOR=#0000cc>e_out_</font>) &#8594; <FONT COLOR=#0000cc>g_out_</font>
<BR><FONT COLOR=#0000cc>U7_</font>(<FONT COLOR=#0000cc>g_out_</font>) &#8594; <FONT COLOR=#0000cc>f_out_</font>
<BR><FONT COLOR=#0000cc>U6_</font>(<FONT COLOR=#0000cc>f_out_</font>) &#8594; <FONT COLOR=#0000cc>e_out_</font>
<BR><FONT COLOR=#0000cc>U2_</font>(<FONT COLOR=#0000cc>e_out_</font>) &#8594; <FONT COLOR=#0000cc>a_out_</font></BLOCKQUOTE><BR>Pi is empty.<BR>We have to consider all (P,R,Pi)-chains<BR>The approximation of the Dependency Graph [30] contains 2 SCCs with 10 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 AND</pre><pre>              &#8627 <B>PiDP</B></pre><pre>                &#8627 UsableRulesProof</pre><pre>              &#8627 PiDP</pre><BR>Pi DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>E_IN_</font> &#8594; <FONT COLOR=#0000cc>F_IN_</font>
<BR><FONT COLOR=#0000cc>F_IN_</font> &#8594; <FONT COLOR=#0000cc>G_IN_</font>
<BR><FONT COLOR=#0000cc>G_IN_</font> &#8594; <FONT COLOR=#0000cc>E_IN_</font></BLOCKQUOTE><BR>The TRS R consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>a_in_</font> &#8594; <FONT COLOR=#0000cc>U1_</font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>b_in_</font> &#8594; <FONT COLOR=#0000cc>U3_</font>(<FONT COLOR=#0000cc>c_in_</font>)
<BR><FONT COLOR=#0000cc>c_in_</font> &#8594; <FONT COLOR=#0000cc>U4_</font>(<FONT COLOR=#0000cc>d_in_</font>)
<BR><FONT COLOR=#0000cc>d_in_</font> &#8594; <FONT COLOR=#0000cc>U5_</font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>U5_</font>(<FONT COLOR=#0000cc>b_out_</font>) &#8594; <FONT COLOR=#0000cc>d_out_</font>
<BR><FONT COLOR=#0000cc>U4_</font>(<FONT COLOR=#0000cc>d_out_</font>) &#8594; <FONT COLOR=#0000cc>c_out_</font>
<BR><FONT COLOR=#0000cc>U3_</font>(<FONT COLOR=#0000cc>c_out_</font>) &#8594; <FONT COLOR=#0000cc>b_out_</font>
<BR><FONT COLOR=#0000cc>U1_</font>(<FONT COLOR=#0000cc>b_out_</font>) &#8594; <FONT COLOR=#0000cc>a_out_</font>
<BR><FONT COLOR=#0000cc>a_in_</font> &#8594; <FONT COLOR=#0000cc>U2_</font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>e_in_</font> &#8594; <FONT COLOR=#0000cc>U6_</font>(<FONT COLOR=#0000cc>f_in_</font>)
<BR><FONT COLOR=#0000cc>f_in_</font> &#8594; <FONT COLOR=#0000cc>U7_</font>(<FONT COLOR=#0000cc>g_in_</font>)
<BR><FONT COLOR=#0000cc>g_in_</font> &#8594; <FONT COLOR=#0000cc>U8_</font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>U8_</font>(<FONT COLOR=#0000cc>e_out_</font>) &#8594; <FONT COLOR=#0000cc>g_out_</font>
<BR><FONT COLOR=#0000cc>U7_</font>(<FONT COLOR=#0000cc>g_out_</font>) &#8594; <FONT COLOR=#0000cc>f_out_</font>
<BR><FONT COLOR=#0000cc>U6_</font>(<FONT COLOR=#0000cc>f_out_</font>) &#8594; <FONT COLOR=#0000cc>e_out_</font>
<BR><FONT COLOR=#0000cc>U2_</font>(<FONT COLOR=#0000cc>e_out_</font>) &#8594; <FONT COLOR=#0000cc>a_out_</font></BLOCKQUOTE><BR>Pi is empty.<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 AND</pre><pre>              &#8627 PiDP</pre><pre>                &#8627 UsableRulesProof</pre><pre>                  &#8627 <B>PiDP</B></pre><pre>                    &#8627 PiDPToQDPProof</pre><pre>              &#8627 PiDP</pre><BR>Pi DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>E_IN_</font> &#8594; <FONT COLOR=#0000cc>F_IN_</font>
<BR><FONT COLOR=#0000cc>F_IN_</font> &#8594; <FONT COLOR=#0000cc>G_IN_</font>
<BR><FONT COLOR=#0000cc>G_IN_</font> &#8594; <FONT COLOR=#0000cc>E_IN_</font></BLOCKQUOTE><BR>R is empty.<BR>Pi is empty.<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 AND</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 NonTerminationProof</pre><pre>              &#8627 PiDP</pre><BR>Q DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>E_IN_</font> &#8594; <FONT COLOR=#0000cc>F_IN_</font>
<BR><FONT COLOR=#0000cc>F_IN_</font> &#8594; <FONT COLOR=#0000cc>G_IN_</font>
<BR><FONT COLOR=#0000cc>G_IN_</font> &#8594; <FONT COLOR=#0000cc>E_IN_</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>E_IN_</font> &#8594; <FONT COLOR=#0000cc>F_IN_</font>
<BR><FONT COLOR=#0000cc>F_IN_</font> &#8594; <FONT COLOR=#0000cc>G_IN_</font>
<BR><FONT COLOR=#0000cc>G_IN_</font> &#8594; <FONT COLOR=#0000cc>E_IN_</font></BLOCKQUOTE><BR>The TRS R consists of the following rules:none<BR><BR><BR>s = <FONT COLOR=#0000cc>F_IN_</font> evaluates to  t =<FONT COLOR=#0000cc>F_IN_</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>F_IN_</font></B> &#8594; <FONT COLOR=#0000cc>G_IN_</font><BR>with rule <FONT COLOR=#0000cc>F_IN_</font> &#8594; <FONT COLOR=#0000cc>G_IN_</font> at position [] and matcher [ ]<BR><BR><B><FONT COLOR=#0000cc>G_IN_</font></B> &#8594; <FONT COLOR=#0000cc>E_IN_</font><BR>with rule <FONT COLOR=#0000cc>G_IN_</font> &#8594; <FONT COLOR=#0000cc>E_IN_</font> at position [] and matcher [ ]<BR><BR><B><FONT COLOR=#0000cc>E_IN_</font></B> &#8594; <FONT COLOR=#0000cc>F_IN_</font><BR>with rule <FONT COLOR=#0000cc>E_IN_</font> &#8594; <FONT COLOR=#0000cc>F_IN_</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><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 AND</pre><pre>              &#8627 PiDP</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>C_IN_</font> &#8594; <FONT COLOR=#0000cc>D_IN_</font>
<BR><FONT COLOR=#0000cc>B_IN_</font> &#8594; <FONT COLOR=#0000cc>C_IN_</font>
<BR><FONT COLOR=#0000cc>D_IN_</font> &#8594; <FONT COLOR=#0000cc>B_IN_</font></BLOCKQUOTE><BR>The TRS R consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>a_in_</font> &#8594; <FONT COLOR=#0000cc>U1_</font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>b_in_</font> &#8594; <FONT COLOR=#0000cc>U3_</font>(<FONT COLOR=#0000cc>c_in_</font>)
<BR><FONT COLOR=#0000cc>c_in_</font> &#8594; <FONT COLOR=#0000cc>U4_</font>(<FONT COLOR=#0000cc>d_in_</font>)
<BR><FONT COLOR=#0000cc>d_in_</font> &#8594; <FONT COLOR=#0000cc>U5_</font>(<FONT COLOR=#0000cc>b_in_</font>)
<BR><FONT COLOR=#0000cc>U5_</font>(<FONT COLOR=#0000cc>b_out_</font>) &#8594; <FONT COLOR=#0000cc>d_out_</font>
<BR><FONT COLOR=#0000cc>U4_</font>(<FONT COLOR=#0000cc>d_out_</font>) &#8594; <FONT COLOR=#0000cc>c_out_</font>
<BR><FONT COLOR=#0000cc>U3_</font>(<FONT COLOR=#0000cc>c_out_</font>) &#8594; <FONT COLOR=#0000cc>b_out_</font>
<BR><FONT COLOR=#0000cc>U1_</font>(<FONT COLOR=#0000cc>b_out_</font>) &#8594; <FONT COLOR=#0000cc>a_out_</font>
<BR><FONT COLOR=#0000cc>a_in_</font> &#8594; <FONT COLOR=#0000cc>U2_</font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>e_in_</font> &#8594; <FONT COLOR=#0000cc>U6_</font>(<FONT COLOR=#0000cc>f_in_</font>)
<BR><FONT COLOR=#0000cc>f_in_</font> &#8594; <FONT COLOR=#0000cc>U7_</font>(<FONT COLOR=#0000cc>g_in_</font>)
<BR><FONT COLOR=#0000cc>g_in_</font> &#8594; <FONT COLOR=#0000cc>U8_</font>(<FONT COLOR=#0000cc>e_in_</font>)
<BR><FONT COLOR=#0000cc>U8_</font>(<FONT COLOR=#0000cc>e_out_</font>) &#8594; <FONT COLOR=#0000cc>g_out_</font>
<BR><FONT COLOR=#0000cc>U7_</font>(<FONT COLOR=#0000cc>g_out_</font>) &#8594; <FONT COLOR=#0000cc>f_out_</font>
<BR><FONT COLOR=#0000cc>U6_</font>(<FONT COLOR=#0000cc>f_out_</font>) &#8594; <FONT COLOR=#0000cc>e_out_</font>
<BR><FONT COLOR=#0000cc>U2_</font>(<FONT COLOR=#0000cc>e_out_</font>) &#8594; <FONT COLOR=#0000cc>a_out_</font></BLOCKQUOTE><BR>Pi is empty.<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 AND</pre><pre>              &#8627 PiDP</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>C_IN_</font> &#8594; <FONT COLOR=#0000cc>D_IN_</font>
<BR><FONT COLOR=#0000cc>B_IN_</font> &#8594; <FONT COLOR=#0000cc>C_IN_</font>
<BR><FONT COLOR=#0000cc>D_IN_</font> &#8594; <FONT COLOR=#0000cc>B_IN_</font></BLOCKQUOTE><BR>R is empty.<BR>Pi is empty.<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 AND</pre><pre>              &#8627 PiDP</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 NonTerminationProof</pre><BR>Q DP problem:<BR>The TRS P consists of the following rules:<BR><BLOCKQUOTE><BR><FONT COLOR=#0000cc>C_IN_</font> &#8594; <FONT COLOR=#0000cc>D_IN_</font>
<BR><FONT COLOR=#0000cc>B_IN_</font> &#8594; <FONT COLOR=#0000cc>C_IN_</font>
<BR><FONT COLOR=#0000cc>D_IN_</font> &#8594; <FONT COLOR=#0000cc>B_IN_</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>C_IN_</font> &#8594; <FONT COLOR=#0000cc>D_IN_</font>
<BR><FONT COLOR=#0000cc>B_IN_</font> &#8594; <FONT COLOR=#0000cc>C_IN_</font>
<BR><FONT COLOR=#0000cc>D_IN_</font> &#8594; <FONT COLOR=#0000cc>B_IN_</font></BLOCKQUOTE><BR>The TRS R consists of the following rules:none<BR><BR><BR>s = <FONT COLOR=#0000cc>D_IN_</font> evaluates to  t =<FONT COLOR=#0000cc>D_IN_</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>D_IN_</font></B> &#8594; <FONT COLOR=#0000cc>B_IN_</font><BR>with rule <FONT COLOR=#0000cc>D_IN_</font> &#8594; <FONT COLOR=#0000cc>B_IN_</font> at position [] and matcher [ ]<BR><BR><B><FONT COLOR=#0000cc>B_IN_</font></B> &#8594; <FONT COLOR=#0000cc>C_IN_</font><BR>with rule <FONT COLOR=#0000cc>B_IN_</font> &#8594; <FONT COLOR=#0000cc>C_IN_</font> at position [] and matcher [ ]<BR><BR><B><FONT COLOR=#0000cc>C_IN_</font></B> &#8594; <FONT COLOR=#0000cc>D_IN_</font><BR>with rule <FONT COLOR=#0000cc>C_IN_</font> &#8594; <FONT COLOR=#0000cc>D_IN_</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>

