(0) Obligation:
Clauses:
p(X) :- ','(q(a), !).
p(X) :- p(X).
q(b).
Queries:
p(a).
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph.
(2) Obligation:
Triples:
p1(T9) :- p1(T9).
Clauses:
pc1(T9) :- pc1(T9).
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(T9) → U1_A(T9, p1_in_a(T9))
P1_IN_A(T9) → P1_IN_A(T9)
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(T9) → U1_A(T9, p1_in_a(T9))
P1_IN_A(T9) → P1_IN_A(T9)
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(T9) → P1_IN_A(T9)
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_AThus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [ ]
- Semiunifier: [ ]
Rewriting sequenceThe DP semiunifies directly so there is only one rewrite step from P1_IN_A to P1_IN_A.
(10) NO