Left Termination of the query pattern
p_in_2(g, a)
w.r.t. the given Prolog program could successfully be proven:
↳ Prolog
↳ PrologToPiTRSProof
Clauses:
p(X, X).
p(f(X), g(Y)) :- ','(p(f(X), f(Z)), p(Z, g(Y))).
Queries:
p(g,a).
We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
p_in(f(X), g(Y)) → U1(X, Y, p_in(f(X), f(Z)))
p_in(X, X) → p_out(X, X)
U1(X, Y, p_out(f(X), f(Z))) → U2(X, Y, Z, p_in(Z, g(Y)))
U2(X, Y, Z, p_out(Z, g(Y))) → p_out(f(X), g(Y))
The argument filtering Pi contains the following mapping:
p_in(x1, x2) = p_in(x1)
f(x1) = f(x1)
g(x1) = g(x1)
U1(x1, x2, x3) = U1(x3)
p_out(x1, x2) = p_out(x2)
U2(x1, x2, x3, x4) = U2(x4)
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:
p_in(f(X), g(Y)) → U1(X, Y, p_in(f(X), f(Z)))
p_in(X, X) → p_out(X, X)
U1(X, Y, p_out(f(X), f(Z))) → U2(X, Y, Z, p_in(Z, g(Y)))
U2(X, Y, Z, p_out(Z, g(Y))) → p_out(f(X), g(Y))
The argument filtering Pi contains the following mapping:
p_in(x1, x2) = p_in(x1)
f(x1) = f(x1)
g(x1) = g(x1)
U1(x1, x2, x3) = U1(x3)
p_out(x1, x2) = p_out(x2)
U2(x1, x2, x3, x4) = U2(x4)
Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:
P_IN(f(X), g(Y)) → U11(X, Y, p_in(f(X), f(Z)))
P_IN(f(X), g(Y)) → P_IN(f(X), f(Z))
U11(X, Y, p_out(f(X), f(Z))) → U21(X, Y, Z, p_in(Z, g(Y)))
U11(X, Y, p_out(f(X), f(Z))) → P_IN(Z, g(Y))
The TRS R consists of the following rules:
p_in(f(X), g(Y)) → U1(X, Y, p_in(f(X), f(Z)))
p_in(X, X) → p_out(X, X)
U1(X, Y, p_out(f(X), f(Z))) → U2(X, Y, Z, p_in(Z, g(Y)))
U2(X, Y, Z, p_out(Z, g(Y))) → p_out(f(X), g(Y))
The argument filtering Pi contains the following mapping:
p_in(x1, x2) = p_in(x1)
f(x1) = f(x1)
g(x1) = g(x1)
U1(x1, x2, x3) = U1(x3)
p_out(x1, x2) = p_out(x2)
U2(x1, x2, x3, x4) = U2(x4)
P_IN(x1, x2) = P_IN(x1)
U21(x1, x2, x3, x4) = U21(x4)
U11(x1, x2, x3) = U11(x3)
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:
P_IN(f(X), g(Y)) → U11(X, Y, p_in(f(X), f(Z)))
P_IN(f(X), g(Y)) → P_IN(f(X), f(Z))
U11(X, Y, p_out(f(X), f(Z))) → U21(X, Y, Z, p_in(Z, g(Y)))
U11(X, Y, p_out(f(X), f(Z))) → P_IN(Z, g(Y))
The TRS R consists of the following rules:
p_in(f(X), g(Y)) → U1(X, Y, p_in(f(X), f(Z)))
p_in(X, X) → p_out(X, X)
U1(X, Y, p_out(f(X), f(Z))) → U2(X, Y, Z, p_in(Z, g(Y)))
U2(X, Y, Z, p_out(Z, g(Y))) → p_out(f(X), g(Y))
The argument filtering Pi contains the following mapping:
p_in(x1, x2) = p_in(x1)
f(x1) = f(x1)
g(x1) = g(x1)
U1(x1, x2, x3) = U1(x3)
p_out(x1, x2) = p_out(x2)
U2(x1, x2, x3, x4) = U2(x4)
P_IN(x1, x2) = P_IN(x1)
U21(x1, x2, x3, x4) = U21(x4)
U11(x1, x2, x3) = U11(x3)
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:
U11(X, Y, p_out(f(X), f(Z))) → P_IN(Z, g(Y))
P_IN(f(X), g(Y)) → U11(X, Y, p_in(f(X), f(Z)))
The TRS R consists of the following rules:
p_in(f(X), g(Y)) → U1(X, Y, p_in(f(X), f(Z)))
p_in(X, X) → p_out(X, X)
U1(X, Y, p_out(f(X), f(Z))) → U2(X, Y, Z, p_in(Z, g(Y)))
U2(X, Y, Z, p_out(Z, g(Y))) → p_out(f(X), g(Y))
The argument filtering Pi contains the following mapping:
p_in(x1, x2) = p_in(x1)
f(x1) = f(x1)
g(x1) = g(x1)
U1(x1, x2, x3) = U1(x3)
p_out(x1, x2) = p_out(x2)
U2(x1, x2, x3, x4) = U2(x4)
P_IN(x1, x2) = P_IN(x1)
U11(x1, x2, x3) = U11(x3)
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:
U11(X, Y, p_out(f(X), f(Z))) → P_IN(Z, g(Y))
P_IN(f(X), g(Y)) → U11(X, Y, p_in(f(X), f(Z)))
The TRS R consists of the following rules:
p_in(X, X) → p_out(X, X)
The argument filtering Pi contains the following mapping:
p_in(x1, x2) = p_in(x1)
f(x1) = f(x1)
g(x1) = g(x1)
p_out(x1, x2) = p_out(x2)
P_IN(x1, x2) = P_IN(x1)
U11(x1, x2, x3) = U11(x3)
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
↳ RuleRemovalProof
Q DP problem:
The TRS P consists of the following rules:
P_IN(f(X)) → U11(p_in(f(X)))
U11(p_out(f(Z))) → P_IN(Z)
The TRS R consists of the following rules:
p_in(X) → p_out(X)
The set Q consists of the following terms:
p_in(x0)
We have to consider all (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
P_IN(f(X)) → U11(p_in(f(X)))
U11(p_out(f(Z))) → P_IN(Z)
Used ordering: POLO with Polynomial interpretation [25]:
POL(P_IN(x1)) = 2 + 2·x1
POL(U11(x1)) = 1 + 2·x1
POL(f(x1)) = 1 + 2·x1
POL(p_in(x1)) = x1
POL(p_out(x1)) = x1
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
p_in(X) → p_out(X)
The set Q consists of the following terms:
p_in(x0)
We have to consider all (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.