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:

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

Queries:

plus(g,a,a).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

(2) Obligation:

Clauses:

plus1(0, T5, T5).
plus1(s(T18), T12, s(T13)) :- plus1(T18, T12, T13).

Queries:

plus1(g,a,a).

(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:
plus1_in: (b,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

plus1_in_gaa(0, T5, T5) → plus1_out_gaa(0, T5, T5)
plus1_in_gaa(s(T18), T12, s(T13)) → U1_gaa(T18, T12, T13, plus1_in_gaa(T18, T12, T13))
U1_gaa(T18, T12, T13, plus1_out_gaa(T18, T12, T13)) → plus1_out_gaa(s(T18), T12, s(T13))

The argument filtering Pi contains the following mapping:
plus1_in_gaa(x1, x2, x3)  =  plus1_in_gaa(x1)
0  =  0
plus1_out_gaa(x1, x2, x3)  =  plus1_out_gaa
s(x1)  =  s(x1)
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x4)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

plus1_in_gaa(0, T5, T5) → plus1_out_gaa(0, T5, T5)
plus1_in_gaa(s(T18), T12, s(T13)) → U1_gaa(T18, T12, T13, plus1_in_gaa(T18, T12, T13))
U1_gaa(T18, T12, T13, plus1_out_gaa(T18, T12, T13)) → plus1_out_gaa(s(T18), T12, s(T13))

The argument filtering Pi contains the following mapping:
plus1_in_gaa(x1, x2, x3)  =  plus1_in_gaa(x1)
0  =  0
plus1_out_gaa(x1, x2, x3)  =  plus1_out_gaa
s(x1)  =  s(x1)
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x4)

(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:

PLUS1_IN_GAA(s(T18), T12, s(T13)) → U1_GAA(T18, T12, T13, plus1_in_gaa(T18, T12, T13))
PLUS1_IN_GAA(s(T18), T12, s(T13)) → PLUS1_IN_GAA(T18, T12, T13)

The TRS R consists of the following rules:

plus1_in_gaa(0, T5, T5) → plus1_out_gaa(0, T5, T5)
plus1_in_gaa(s(T18), T12, s(T13)) → U1_gaa(T18, T12, T13, plus1_in_gaa(T18, T12, T13))
U1_gaa(T18, T12, T13, plus1_out_gaa(T18, T12, T13)) → plus1_out_gaa(s(T18), T12, s(T13))

The argument filtering Pi contains the following mapping:
plus1_in_gaa(x1, x2, x3)  =  plus1_in_gaa(x1)
0  =  0
plus1_out_gaa(x1, x2, x3)  =  plus1_out_gaa
s(x1)  =  s(x1)
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x4)
PLUS1_IN_GAA(x1, x2, x3)  =  PLUS1_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x4)

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

(6) Obligation:

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

PLUS1_IN_GAA(s(T18), T12, s(T13)) → U1_GAA(T18, T12, T13, plus1_in_gaa(T18, T12, T13))
PLUS1_IN_GAA(s(T18), T12, s(T13)) → PLUS1_IN_GAA(T18, T12, T13)

The TRS R consists of the following rules:

plus1_in_gaa(0, T5, T5) → plus1_out_gaa(0, T5, T5)
plus1_in_gaa(s(T18), T12, s(T13)) → U1_gaa(T18, T12, T13, plus1_in_gaa(T18, T12, T13))
U1_gaa(T18, T12, T13, plus1_out_gaa(T18, T12, T13)) → plus1_out_gaa(s(T18), T12, s(T13))

The argument filtering Pi contains the following mapping:
plus1_in_gaa(x1, x2, x3)  =  plus1_in_gaa(x1)
0  =  0
plus1_out_gaa(x1, x2, x3)  =  plus1_out_gaa
s(x1)  =  s(x1)
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x4)
PLUS1_IN_GAA(x1, x2, x3)  =  PLUS1_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x4)

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:

PLUS1_IN_GAA(s(T18), T12, s(T13)) → PLUS1_IN_GAA(T18, T12, T13)

The TRS R consists of the following rules:

plus1_in_gaa(0, T5, T5) → plus1_out_gaa(0, T5, T5)
plus1_in_gaa(s(T18), T12, s(T13)) → U1_gaa(T18, T12, T13, plus1_in_gaa(T18, T12, T13))
U1_gaa(T18, T12, T13, plus1_out_gaa(T18, T12, T13)) → plus1_out_gaa(s(T18), T12, s(T13))

The argument filtering Pi contains the following mapping:
plus1_in_gaa(x1, x2, x3)  =  plus1_in_gaa(x1)
0  =  0
plus1_out_gaa(x1, x2, x3)  =  plus1_out_gaa
s(x1)  =  s(x1)
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x4)
PLUS1_IN_GAA(x1, x2, x3)  =  PLUS1_IN_GAA(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:

PLUS1_IN_GAA(s(T18), T12, s(T13)) → PLUS1_IN_GAA(T18, T12, T13)

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

(11) PiDPToQDPProof (SOUND 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:

PLUS1_IN_GAA(s(T18)) → PLUS1_IN_GAA(T18)

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:

• PLUS1_IN_GAA(s(T18)) → PLUS1_IN_GAA(T18)
  The graph contains the following edges 1 > 1

(14) YES