(0) Obligation:
Clauses:p(X) :- ','(q(f(Y)), p(Y)).
p(g(X)) :- p(X).
q(g(Y)).
Queries:
p(g).
(1) PrologToPrologProblemTransformerProof (SOUND transformation)
Built Prolog problem from termination graph.
(2) Obligation:
Clauses:p1(g(T6)) :- p1(T6).
Queries:
p1(g).
(3) PrologToPiTRSProof (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 Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
p1_in_g(g(T6)) → U1_g(T6, p1_in_g(T6))
U1_g(T6, p1_out_g(T6)) → p1_out_g(g(T6))
The argument filtering Pi contains the following mapping:
p1_in_g(x1) = p1_in_g(x1)
g(x1) = g(x1)
U1_g(x1, x2) = U1_g(x2)
p1_out_g(x1) = p1_out_g
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(4) Obligation:
Pi-finite rewrite system:The TRS R consists of the following rules:
p1_in_g(g(T6)) → U1_g(T6, p1_in_g(T6))
U1_g(T6, p1_out_g(T6)) → p1_out_g(g(T6))
The argument filtering Pi contains the following mapping:
p1_in_g(x1) = p1_in_g(x1)
g(x1) = g(x1)
U1_g(x1, x2) = U1_g(x2)
p1_out_g(x1) = p1_out_g
(5) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem: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)
The TRS R consists of the following rules:
p1_in_g(g(T6)) → U1_g(T6, p1_in_g(T6))
U1_g(T6, p1_out_g(T6)) → p1_out_g(g(T6))
The argument filtering Pi contains the following mapping:
p1_in_g(x1) = p1_in_g(x1)
g(x1) = g(x1)
U1_g(x1, x2) = U1_g(x2)
p1_out_g(x1) = p1_out_g
P1_IN_G(x1) = P1_IN_G(x1)
U1_G(x1, x2) = U1_G(x2)
We have to consider all (P,R,Pi)-chains
(6) 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)
The TRS R consists of the following rules:
p1_in_g(g(T6)) → U1_g(T6, p1_in_g(T6))
U1_g(T6, p1_out_g(T6)) → p1_out_g(g(T6))
The argument filtering Pi contains the following mapping:
p1_in_g(x1) = p1_in_g(x1)
g(x1) = g(x1)
U1_g(x1, x2) = U1_g(x2)
p1_out_g(x1) = p1_out_g
P1_IN_G(x1) = P1_IN_G(x1)
U1_G(x1, x2) = U1_G(x2)
We have to consider all (P,R,Pi)-chains
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.(8) Obligation:
Pi DP problem:The TRS P consists of the following rules:
P1_IN_G(g(T6)) → P1_IN_G(T6)
The TRS R consists of the following rules:
p1_in_g(g(T6)) → U1_g(T6, p1_in_g(T6))
U1_g(T6, p1_out_g(T6)) → p1_out_g(g(T6))
The argument filtering Pi contains the following mapping:
p1_in_g(x1) = p1_in_g(x1)
g(x1) = g(x1)
U1_g(x1, x2) = U1_g(x2)
p1_out_g(x1) = p1_out_g
P1_IN_G(x1) = P1_IN_G(x1)
We have to consider all (P,R,Pi)-chains
(9) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.(10) 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
(11) PiDPToQDPProof (EQUIVALENT transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.(12) 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.
(13) 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