(0) Obligation:

Clauses:

q(X) :- ','(not_zero(X), ','(p(X, Y), q(Y))).
p(0, 0).
p(s(X), X).
zero(0).
not_zero(X) :- ','(zero(X), ','(!, failure(a))).
not_zero(X1).
failure(b).

Queries:

q(g).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

(2) Obligation:

Clauses:

q1(s(T5)) :- q1(T5).

Queries:

q1(g).

(3) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
q1_in: (b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

q1_in_g(s(T5)) → U1_g(T5, q1_in_g(T5))
U1_g(T5, q1_out_g(T5)) → q1_out_g(s(T5))

The argument filtering Pi contains the following mapping:
q1_in_g(x1)  =  q1_in_g(x1)
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)
q1_out_g(x1)  =  q1_out_g

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

Pi-finite rewrite system:
The TRS R consists of the following rules:

q1_in_g(s(T5)) → U1_g(T5, q1_in_g(T5))
U1_g(T5, q1_out_g(T5)) → q1_out_g(s(T5))

The argument filtering Pi contains the following mapping:
q1_in_g(x1)  =  q1_in_g(x1)
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)
q1_out_g(x1)  =  q1_out_g

(5) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

Q1_IN_G(s(T5)) → U1_G(T5, q1_in_g(T5))
Q1_IN_G(s(T5)) → Q1_IN_G(T5)

The TRS R consists of the following rules:

q1_in_g(s(T5)) → U1_g(T5, q1_in_g(T5))
U1_g(T5, q1_out_g(T5)) → q1_out_g(s(T5))

The argument filtering Pi contains the following mapping:
q1_in_g(x1)  =  q1_in_g(x1)
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)
q1_out_g(x1)  =  q1_out_g
Q1_IN_G(x1)  =  Q1_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)

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

(6) Obligation:

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

Q1_IN_G(s(T5)) → U1_G(T5, q1_in_g(T5))
Q1_IN_G(s(T5)) → Q1_IN_G(T5)

The TRS R consists of the following rules:

q1_in_g(s(T5)) → U1_g(T5, q1_in_g(T5))
U1_g(T5, q1_out_g(T5)) → q1_out_g(s(T5))

The argument filtering Pi contains the following mapping:
q1_in_g(x1)  =  q1_in_g(x1)
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)
q1_out_g(x1)  =  q1_out_g
Q1_IN_G(x1)  =  Q1_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.

(8) Obligation:

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

Q1_IN_G(s(T5)) → Q1_IN_G(T5)

The TRS R consists of the following rules:

q1_in_g(s(T5)) → U1_g(T5, q1_in_g(T5))
U1_g(T5, q1_out_g(T5)) → q1_out_g(s(T5))

The argument filtering Pi contains the following mapping:
q1_in_g(x1)  =  q1_in_g(x1)
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)
q1_out_g(x1)  =  q1_out_g
Q1_IN_G(x1)  =  Q1_IN_G(x1)

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

(9) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(10) Obligation:

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

Q1_IN_G(s(T5)) → Q1_IN_G(T5)

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

(11) PiDPToQDPProof (EQUIVALENT transformation)

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

(12) Obligation:

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

Q1_IN_G(s(T5)) → Q1_IN_G(T5)

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

(13) 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:

  • Q1_IN_G(s(T5)) → Q1_IN_G(T5)
    The graph contains the following edges 1 > 1

(14) TRUE