(0) Obligation:

Clauses:

num(0) :- !.
num(X) :- ','(p(X, Y), num(Y)).
p(0, 0).
p(s(X), X).

Queries:

num(g).

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(2) Obligation:

Triples:

num1(s(T6)) :- num1(T6).

Clauses:

numc1(0).
numc1(s(T6)) :- numc1(T6).

Afs:

num1(x1)  =  num1(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:
num1_in: (b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

NUM1_IN_G(s(T6)) → U1_G(T6, num1_in_g(T6))
NUM1_IN_G(s(T6)) → NUM1_IN_G(T6)

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:

NUM1_IN_G(s(T6)) → U1_G(T6, num1_in_g(T6))
NUM1_IN_G(s(T6)) → NUM1_IN_G(T6)

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 1 less node.

(6) Obligation:

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

NUM1_IN_G(s(T6)) → NUM1_IN_G(T6)

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:

NUM1_IN_G(s(T6)) → NUM1_IN_G(T6)

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

(9) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • NUM1_IN_G(s(T6)) → NUM1_IN_G(T6)
    The graph contains the following edges 1 > 1

(10) YES