(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_AP1_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.



(10) NO