(0) Obligation:
Clauses:p(X) :- q(X).
p(X) :- p(X).
q(X) :- !.
Queries:
p(a).
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph.
(2) Obligation:
Triples:p1(T13) :- p1(T13).
Clauses:
pc1(T7).
pc1(T13) :- pc1(T13).
Afs:
p1(x1) = p1
(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: (f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
P1_IN_A(T13) → U1_A(T13, p1_in_a(T13))
P1_IN_A(T13) → P1_IN_A(T13)
R is empty.
The argument filtering Pi contains the following mapping:
p1_in_a(x1) = p1_in_a
P1_IN_A(x1) = P1_IN_A
U1_A(x1, x2) = U1_A(x2)
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_A(T13) → U1_A(T13, p1_in_a(T13))
P1_IN_A(T13) → P1_IN_A(T13)
R is empty.
The argument filtering Pi contains the following mapping:
p1_in_a(x1) = p1_in_a
P1_IN_A(x1) = P1_IN_A
U1_A(x1, x2) = U1_A(x2)
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_A(T13) → P1_IN_A(T13)
R is empty.
The argument filtering Pi contains the following mapping:
P1_IN_A(x1) = P1_IN_A
We have to consider all (P,R,Pi)-chains
(7) PiDPToQDPProof (SOUND 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_A → P1_IN_A
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(9) NonTerminationProof (EQUIVALENT transformation)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.Found a loop by semiunifying a rule from P directly.
s = P1_IN_A evaluates to t =P1_IN_A
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [ ]
Rewriting sequence
The DP semiunifies directly so there is only one rewrite step from P1_IN_A to P1_IN_A.