Left Termination of the query pattern plus_in_3(g, a, a) w.r.t. the given Prolog program could successfully be proven:

(0) Obligation:

Clauses:

plus(0, Y, Y).
plus(s(X), Y, Z) :- plus(X, s(Y), Z).

Queries:

plus(g,a,a).

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(2) Obligation:

Triples:

plus1(s(s(T25)), T29, T28) :- plus1(T25, s(s(T29)), T28).

Clauses:

plusc1(0, T5, T5).
plusc1(s(0), T18, s(T18)).
plusc1(s(s(T25)), T29, T28) :- plusc1(T25, s(s(T29)), T28).

Afs:

plus1(x1, x2, x3)  =  plus1(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:
plus1_in: (b,f,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

PLUS1_IN_GAA(s(s(T25)), T29, T28) → U1_GAA(T25, T29, T28, plus1_in_gaa(T25, s(s(T29)), T28))
PLUS1_IN_GAA(s(s(T25)), T29, T28) → PLUS1_IN_GAA(T25, s(s(T29)), T28)

R is empty.
The argument filtering Pi contains the following mapping:
plus1_in_gaa(x1, x2, x3)  =  plus1_in_gaa(x1)
s(x1)  =  s(x1)
PLUS1_IN_GAA(x1, x2, x3)  =  PLUS1_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x1, 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:

PLUS1_IN_GAA(s(s(T25)), T29, T28) → U1_GAA(T25, T29, T28, plus1_in_gaa(T25, s(s(T29)), T28))
PLUS1_IN_GAA(s(s(T25)), T29, T28) → PLUS1_IN_GAA(T25, s(s(T29)), T28)

R is empty.
The argument filtering Pi contains the following mapping:
plus1_in_gaa(x1, x2, x3)  =  plus1_in_gaa(x1)
s(x1)  =  s(x1)
PLUS1_IN_GAA(x1, x2, x3)  =  PLUS1_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x1, 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:

PLUS1_IN_GAA(s(s(T25)), T29, T28) → PLUS1_IN_GAA(T25, s(s(T29)), T28)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
PLUS1_IN_GAA(x1, x2, x3)  =  PLUS1_IN_GAA(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:

PLUS1_IN_GAA(s(s(T25))) → PLUS1_IN_GAA(T25)

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:

• PLUS1_IN_GAA(s(s(T25))) → PLUS1_IN_GAA(T25)
  The graph contains the following edges 1 > 1

(10) YES