(0) Obligation:
Clauses:p(X) :- ','(q(f(Y)), p(Y)).
p(g(X)) :- p(X).
q(g(Y)).
Queries:
p(g).
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph.
(2) Obligation:
Triples:p1(g(T6)) :- p1(T6).
Clauses:
pc1(g(T6)) :- pc1(T6).
Afs:
p1(x1) = p1(x1)
(3) TriplesToPiDPProof (SOUND transformation)
We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:p1_in: (b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
P1_IN_G(g(T6)) → U1_G(T6, p1_in_g(T6))
P1_IN_G(g(T6)) → P1_IN_G(T6)
R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains
Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES
(4) Obligation:
Pi DP problem:The TRS P consists of the following rules:
P1_IN_G(g(T6)) → U1_G(T6, p1_in_g(T6))
P1_IN_G(g(T6)) → P1_IN_G(T6)
R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.(6) Obligation:
Pi DP problem:The TRS P consists of the following rules:
P1_IN_G(g(T6)) → P1_IN_G(T6)
R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains
(7) PiDPToQDPProof (EQUIVALENT transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.(8) Obligation:
Q DP problem:The TRS P consists of the following rules:
P1_IN_G(g(T6)) → P1_IN_G(T6)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(9) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.From the DPs we obtained the following set of size-change graphs:
- P1_IN_G(g(T6)) → P1_IN_G(T6)
The graph contains the following edges 1 > 1