Left Termination of the query pattern add_in_3(a, g, a) w.r.t. the given Prolog program could successfully be proven:

(0) Obligation:

Clauses:

add(X, 0, X).
add(X, Y, s(Z)) :- ','(no(zero(Y)), ','(p(Y, P), add(X, P, Z))).
p(0, 0).
p(s(X), X).
zero(0).
no(X) :- ','(X, ','(!, failure(a))).
no(X).
failure(b).

Queries:

add(a,g,a).

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(2) Obligation:

Triples:

add1(T20, s(T36), s(T21)) :- add1(T20, T36, T21).

Clauses:

addc1(T5, 0, T5).
addc1(T20, s(T36), s(T21)) :- addc1(T20, T36, T21).

Afs:

add1(x1, x2, x3)  =  add1(x2)

(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:
add1_in: (f,b,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

ADD1_IN_AGA(T20, s(T36), s(T21)) → U1_AGA(T20, T36, T21, add1_in_aga(T20, T36, T21))
ADD1_IN_AGA(T20, s(T36), s(T21)) → ADD1_IN_AGA(T20, T36, T21)

R is empty.
The argument filtering Pi contains the following mapping:
add1_in_aga(x1, x2, x3)  =  add1_in_aga(x2)
s(x1)  =  s(x1)
ADD1_IN_AGA(x1, x2, x3)  =  ADD1_IN_AGA(x2)
U1_AGA(x1, x2, x3, x4)  =  U1_AGA(x2, x4)

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:

ADD1_IN_AGA(T20, s(T36), s(T21)) → U1_AGA(T20, T36, T21, add1_in_aga(T20, T36, T21))
ADD1_IN_AGA(T20, s(T36), s(T21)) → ADD1_IN_AGA(T20, T36, T21)

R is empty.
The argument filtering Pi contains the following mapping:
add1_in_aga(x1, x2, x3)  =  add1_in_aga(x2)
s(x1)  =  s(x1)
ADD1_IN_AGA(x1, x2, x3)  =  ADD1_IN_AGA(x2)
U1_AGA(x1, x2, x3, x4)  =  U1_AGA(x2, x4)

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:

ADD1_IN_AGA(T20, s(T36), s(T21)) → ADD1_IN_AGA(T20, T36, T21)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
ADD1_IN_AGA(x1, x2, x3)  =  ADD1_IN_AGA(x2)

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:

ADD1_IN_AGA(s(T36)) → ADD1_IN_AGA(T36)

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:

• ADD1_IN_AGA(s(T36)) → ADD1_IN_AGA(T36)
  The graph contains the following edges 1 > 1

(10) YES