(0) Obligation:
Clauses:q(X) :- ','(p(X), !).
p(0).
p(s(X)) :- p(X).
Queries:
q(a).
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph.
(2) Obligation:
Triples:p3(s(T8)) :- p3(T8).
q1(T4) :- p3(T4).
Clauses:
qc3(0).
qc3(s(T8)) :- qc3(T8).
Afs:
q1(x1) = q1
(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:q1_in: (f)
p3_in: (f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
Q1_IN_A(T4) → U2_A(T4, p3_in_a(T4))
Q1_IN_A(T4) → P3_IN_A(T4)
P3_IN_A(s(T8)) → U1_A(T8, p3_in_a(T8))
P3_IN_A(s(T8)) → P3_IN_A(T8)
R is empty.
The argument filtering Pi contains the following mapping:
p3_in_a(x1) = p3_in_a
s(x1) = s(x1)
Q1_IN_A(x1) = Q1_IN_A
U2_A(x1, x2) = U2_A(x2)
P3_IN_A(x1) = P3_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:
Q1_IN_A(T4) → U2_A(T4, p3_in_a(T4))
Q1_IN_A(T4) → P3_IN_A(T4)
P3_IN_A(s(T8)) → U1_A(T8, p3_in_a(T8))
P3_IN_A(s(T8)) → P3_IN_A(T8)
R is empty.
The argument filtering Pi contains the following mapping:
p3_in_a(x1) = p3_in_a
s(x1) = s(x1)
Q1_IN_A(x1) = Q1_IN_A
U2_A(x1, x2) = U2_A(x2)
P3_IN_A(x1) = P3_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 3 less nodes.(6) Obligation:
Pi DP problem:The TRS P consists of the following rules:
P3_IN_A(s(T8)) → P3_IN_A(T8)
R is empty.
The argument filtering Pi contains the following mapping:
s(x1) = s(x1)
P3_IN_A(x1) = P3_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:
P3_IN_A → P3_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 = P3_IN_A evaluates to t =P3_IN_A
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [ ]
- Semiunifier: [ ]
Rewriting sequence
The DP semiunifies directly so there is only one rewrite step from P3_IN_A to P3_IN_A.