Left Termination of the query pattern p_in_1(g) w.r.t. the given Prolog program could successfully be proven:

(0) Obligation:

Clauses:

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

Queries:

p(g).

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(2) Obligation:

Triples:

p7(T8, T8) :- p1(T8).
p7(s(T11), X19) :- p7(T11, X19).
p1(s(T3)) :- p7(T3, X4).

Clauses:

pc1(0).
pc1(s(T3)) :- qc7(T3, X4).
qc7(T8, T8) :- pc1(T8).
qc7(s(T11), X19) :- qc7(T11, X19).

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)
p7_in: (b,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

P1_IN_G(s(T3)) → U3_G(T3, p7_in_ga(T3, X4))
P1_IN_G(s(T3)) → P7_IN_GA(T3, X4)
P7_IN_GA(T8, T8) → U1_GA(T8, p1_in_g(T8))
P7_IN_GA(T8, T8) → P1_IN_G(T8)
P7_IN_GA(s(T11), X19) → U2_GA(T11, X19, p7_in_ga(T11, X19))
P7_IN_GA(s(T11), X19) → P7_IN_GA(T11, X19)

R is empty.
The argument filtering Pi contains the following mapping:
p1_in_g(x1)  =  p1_in_g(x1)
s(x1)  =  s(x1)
p7_in_ga(x1, x2)  =  p7_in_ga(x1)
P1_IN_G(x1)  =  P1_IN_G(x1)
U3_G(x1, x2)  =  U3_G(x1, x2)
P7_IN_GA(x1, x2)  =  P7_IN_GA(x1)
U1_GA(x1, x2)  =  U1_GA(x1, x2)
U2_GA(x1, x2, x3)  =  U2_GA(x1, x3)

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(s(T3)) → U3_G(T3, p7_in_ga(T3, X4))
P1_IN_G(s(T3)) → P7_IN_GA(T3, X4)
P7_IN_GA(T8, T8) → U1_GA(T8, p1_in_g(T8))
P7_IN_GA(T8, T8) → P1_IN_G(T8)
P7_IN_GA(s(T11), X19) → U2_GA(T11, X19, p7_in_ga(T11, X19))
P7_IN_GA(s(T11), X19) → P7_IN_GA(T11, X19)

R is empty.
The argument filtering Pi contains the following mapping:
p1_in_g(x1)  =  p1_in_g(x1)
s(x1)  =  s(x1)
p7_in_ga(x1, x2)  =  p7_in_ga(x1)
P1_IN_G(x1)  =  P1_IN_G(x1)
U3_G(x1, x2)  =  U3_G(x1, x2)
P7_IN_GA(x1, x2)  =  P7_IN_GA(x1)
U1_GA(x1, x2)  =  U1_GA(x1, x2)
U2_GA(x1, x2, x3)  =  U2_GA(x1, x3)

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:

P1_IN_G(s(T3)) → P7_IN_GA(T3, X4)
P7_IN_GA(T8, T8) → P1_IN_G(T8)
P7_IN_GA(s(T11), X19) → P7_IN_GA(T11, X19)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
P1_IN_G(x1)  =  P1_IN_G(x1)
P7_IN_GA(x1, x2)  =  P7_IN_GA(x1)

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_G(s(T3)) → P7_IN_GA(T3)
P7_IN_GA(T8) → P1_IN_G(T8)
P7_IN_GA(s(T11)) → P7_IN_GA(T11)

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:

• P7_IN_GA(T8) → P1_IN_G(T8)
  The graph contains the following edges 1 >= 1

• P7_IN_GA(s(T11)) → P7_IN_GA(T11)
  The graph contains the following edges 1 > 1

• P1_IN_G(s(T3)) → P7_IN_GA(T3)
  The graph contains the following edges 1 > 1

(10) YES