Left Termination of the query pattern p_in_1(g) w.r.t. the given Prolog program could not be shown:

0 Prolog
↳1 PrologToDTProblemTransformerProof (⇐)
↳2 TRIPLES
↳3 TriplesToPiDPProof (⇐)
↳4 PiDP
↳5 DependencyGraphProof (⇔)
↳6 AND
    ↳7 PiDP
        ↳8 PiDPToQDPProof (⇔)
        ↳9 QDP
        ↳10 NonTerminationProof (⇔)
        ↳11 NO
    ↳12 PiDP
        ↳13 PiDPToQDPProof (⇔)
        ↳14 QDP
        ↳15 NonTerminationProof (⇔)
        ↳16 NO

(0) Obligation:

Clauses:

p(a).
p(X) :- p(a).
q(b).

Queries:

p(g).

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(2) Obligation:

Triples:

p6 :- p6.
p13 :- p13.
p1(a) :- p6.
p1(T3) :- p13.

Clauses:

pc6.
pc6 :- pc6.
pc13.
pc13 :- pc13.

Afs:

p1(x1)  =  p1(x1)

(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: (b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

P1_IN_G(a) → U3_G(p6_in_)
P1_IN_G(a) → P6_IN_
P6_IN_ → U1_1(p6_in_)
P6_IN_ → P6_IN_
P1_IN_G(T3) → U4_G(T3, p13_in_)
P1_IN_G(T3) → P13_IN_
P13_IN_ → U2_1(p13_in_)
P13_IN_ → P13_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_G(a) → U3_G(p6_in_)
P1_IN_G(a) → P6_IN_
P6_IN_ → U1_1(p6_in_)
P6_IN_ → P6_IN_
P1_IN_G(T3) → U4_G(T3, p13_in_)
P1_IN_G(T3) → P13_IN_
P13_IN_ → U2_1(p13_in_)
P13_IN_ → P13_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 2 SCCs with 6 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

P13_IN_ → P13_IN_

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

(8) PiDPToQDPProof (EQUIVALENT transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(9) Obligation:

Q DP problem:
The TRS P consists of the following rules:

P13_IN_ → P13_IN_

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

(10) 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 = P13_IN_ evaluates to t =P13_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 P13_IN_ to P13_IN_.



(11) NO

(12) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

P6_IN_ → P6_IN_

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

(13) PiDPToQDPProof (EQUIVALENT transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(14) Obligation:

Q DP problem:
The TRS P consists of the following rules:

P6_IN_ → P6_IN_

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

(15) 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 = P6_IN_ evaluates to t =P6_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 P6_IN_ to P6_IN_.



(16) NO