YES
Left Termination proof of ../tpdb/LP/SGST06/pplus.pl
Left Termination of the query pattern
plus_in_3(g, a, a)
w.r.t. the given Prolog program could successfully be proven:
↳ Prolog
↳ PrologToPiTRSProof
Clauses:
p(s(X), X).
plus(0, Y, Y).
plus(s(X), Y, s(Z)) :- ','(p(s(X), U), plus(U, Y, Z)).
Queries:
plus(g,a,a).
We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
plus_in: (b,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U1_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U1_gaa(X, Y, Z, p_out_ga(s(X), U)) → U2_gaa(X, Y, Z, U, plus_in_gaa(U, Y, Z))
U2_gaa(X, Y, Z, U, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))
The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3) = plus_in_gaa(x1)
0 = 0
plus_out_gaa(x1, x2, x3) = plus_out_gaa
s(x1) = s(x1)
U1_gaa(x1, x2, x3, x4) = U1_gaa(x4)
p_in_ga(x1, x2) = p_in_ga(x1)
p_out_ga(x1, x2) = p_out_ga(x2)
U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
Pi-finite rewrite system:
The TRS R consists of the following rules:
plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U1_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U1_gaa(X, Y, Z, p_out_ga(s(X), U)) → U2_gaa(X, Y, Z, U, plus_in_gaa(U, Y, Z))
U2_gaa(X, Y, Z, U, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))
The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3) = plus_in_gaa(x1)
0 = 0
plus_out_gaa(x1, x2, x3) = plus_out_gaa
s(x1) = s(x1)
U1_gaa(x1, x2, x3, x4) = U1_gaa(x4)
p_in_ga(x1, x2) = p_in_ga(x1)
p_out_ga(x1, x2) = p_out_ga(x2)
U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5)
Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:
PLUS_IN_GAA(s(X), Y, s(Z)) → U1_GAA(X, Y, Z, p_in_ga(s(X), U))
PLUS_IN_GAA(s(X), Y, s(Z)) → P_IN_GA(s(X), U)
U1_GAA(X, Y, Z, p_out_ga(s(X), U)) → U2_GAA(X, Y, Z, U, plus_in_gaa(U, Y, Z))
U1_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)
The TRS R consists of the following rules:
plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U1_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U1_gaa(X, Y, Z, p_out_ga(s(X), U)) → U2_gaa(X, Y, Z, U, plus_in_gaa(U, Y, Z))
U2_gaa(X, Y, Z, U, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))
The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3) = plus_in_gaa(x1)
0 = 0
plus_out_gaa(x1, x2, x3) = plus_out_gaa
s(x1) = s(x1)
U1_gaa(x1, x2, x3, x4) = U1_gaa(x4)
p_in_ga(x1, x2) = p_in_ga(x1)
p_out_ga(x1, x2) = p_out_ga(x2)
U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5)
P_IN_GA(x1, x2) = P_IN_GA(x1)
U2_GAA(x1, x2, x3, x4, x5) = U2_GAA(x5)
U1_GAA(x1, x2, x3, x4) = U1_GAA(x4)
PLUS_IN_GAA(x1, x2, x3) = PLUS_IN_GAA(x1)
We have to consider all (P,R,Pi)-chains
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
Pi DP problem:
The TRS P consists of the following rules:
PLUS_IN_GAA(s(X), Y, s(Z)) → U1_GAA(X, Y, Z, p_in_ga(s(X), U))
PLUS_IN_GAA(s(X), Y, s(Z)) → P_IN_GA(s(X), U)
U1_GAA(X, Y, Z, p_out_ga(s(X), U)) → U2_GAA(X, Y, Z, U, plus_in_gaa(U, Y, Z))
U1_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)
The TRS R consists of the following rules:
plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U1_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U1_gaa(X, Y, Z, p_out_ga(s(X), U)) → U2_gaa(X, Y, Z, U, plus_in_gaa(U, Y, Z))
U2_gaa(X, Y, Z, U, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))
The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3) = plus_in_gaa(x1)
0 = 0
plus_out_gaa(x1, x2, x3) = plus_out_gaa
s(x1) = s(x1)
U1_gaa(x1, x2, x3, x4) = U1_gaa(x4)
p_in_ga(x1, x2) = p_in_ga(x1)
p_out_ga(x1, x2) = p_out_ga(x2)
U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5)
P_IN_GA(x1, x2) = P_IN_GA(x1)
U2_GAA(x1, x2, x3, x4, x5) = U2_GAA(x5)
U1_GAA(x1, x2, x3, x4) = U1_GAA(x4)
PLUS_IN_GAA(x1, x2, x3) = PLUS_IN_GAA(x1)
We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 2 less nodes.
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
Pi DP problem:
The TRS P consists of the following rules:
U1_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)
PLUS_IN_GAA(s(X), Y, s(Z)) → U1_GAA(X, Y, Z, p_in_ga(s(X), U))
The TRS R consists of the following rules:
plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U1_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U1_gaa(X, Y, Z, p_out_ga(s(X), U)) → U2_gaa(X, Y, Z, U, plus_in_gaa(U, Y, Z))
U2_gaa(X, Y, Z, U, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))
The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3) = plus_in_gaa(x1)
0 = 0
plus_out_gaa(x1, x2, x3) = plus_out_gaa
s(x1) = s(x1)
U1_gaa(x1, x2, x3, x4) = U1_gaa(x4)
p_in_ga(x1, x2) = p_in_ga(x1)
p_out_ga(x1, x2) = p_out_ga(x2)
U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5)
U1_GAA(x1, x2, x3, x4) = U1_GAA(x4)
PLUS_IN_GAA(x1, x2, x3) = PLUS_IN_GAA(x1)
We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
Pi DP problem:
The TRS P consists of the following rules:
U1_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)
PLUS_IN_GAA(s(X), Y, s(Z)) → U1_GAA(X, Y, Z, p_in_ga(s(X), U))
The TRS R consists of the following rules:
p_in_ga(s(X), X) → p_out_ga(s(X), X)
The argument filtering Pi contains the following mapping:
s(x1) = s(x1)
p_in_ga(x1, x2) = p_in_ga(x1)
p_out_ga(x1, x2) = p_out_ga(x2)
U1_GAA(x1, x2, x3, x4) = U1_GAA(x4)
PLUS_IN_GAA(x1, x2, x3) = PLUS_IN_GAA(x1)
We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ QDP
↳ RFCMatchBoundsDPProof
Q DP problem:
The TRS P consists of the following rules:
PLUS_IN_GAA(s(X)) → U1_GAA(p_in_ga(s(X)))
U1_GAA(p_out_ga(U)) → PLUS_IN_GAA(U)
The TRS R consists of the following rules:
p_in_ga(s(X)) → p_out_ga(X)
The set Q consists of the following terms:
p_in_ga(x0)
We have to consider all (P,Q,R)-chains.
Termination of the TRS P cup R can be shown by a matchbound [6,7] of 2. This implies finiteness of the given DP problem.
The following rules (P cup R) were used to construct the certificate:
p_in_ga(s(X)) → p_out_ga(X)
PLUS_IN_GAA(s(X)) → U1_GAA(p_in_ga(s(X)))
U1_GAA(p_out_ga(U)) → PLUS_IN_GAA(U)
The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
66, 67, 68, 69, 70, 71
Node 66 is start node and node 67 is final node.
Those nodes are connect through the following edges:
- 66 to 68 labelled U1_GAA_1(0)
- 66 to 67 labelled p_out_ga_1(0), PLUS_IN_GAA_1(0), PLUS_IN_GAA_1(1), PLUS_IN_GAA_1(2)
- 66 to 70 labelled U1_GAA_1(1)
- 67 to 67 labelled #_1(0)
- 68 to 69 labelled p_in_ga_1(0)
- 68 to 67 labelled p_out_ga_1(1)
- 69 to 67 labelled s_1(0)
- 70 to 71 labelled p_in_ga_1(1)
- 70 to 67 labelled p_out_ga_1(2)
- 71 to 67 labelled s_1(1)