Left Termination of the query pattern w.r.t. the given Prolog program could not be shown:

0 Prolog
↳1 PrologToDTProblemTransformerProof (⇐)
↳2 TRIPLES
↳3 TriplesToPiDPProof (⇐)
↳4 PiDP
↳5 DependencyGraphProof (⇔)
↳6 PiDP
↳7 PiDPToQDPProof (⇔)
↳8 QDP
↳9 NonTerminationProof (⇔)
↳10 NO

(0) Obligation:

Clauses:

p :- r.
r :- ','(!, q).
r :- q.
q.
q :- r.

Queries:

p().

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(2) Obligation:

Triples:

r3 :- r3.
p1 :- r3.

Clauses:

rc3.
rc3 :- rc3.

Afs:

p1  =  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:
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

P1_IN_ → U2_1(r3_in_)
P1_IN_ → R3_IN_
R3_IN_ → U1_1(r3_in_)
R3_IN_ → R3_IN_

R is empty.
Pi is empty.
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_ → U2_1(r3_in_)
P1_IN_ → R3_IN_
R3_IN_ → U1_1(r3_in_)
R3_IN_ → R3_IN_

R is empty.
Pi is empty.
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:

R3_IN_ → R3_IN_

R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains

(7) PiDPToQDPProof (EQUIVALENT 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:

R3_IN_ → R3_IN_

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 = R3_IN_ evaluates to t =R3_IN_

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 R3_IN_ to R3_IN_.



(10) NO