Left Termination of the query pattern sum_in_3(a, a, g) w.r.t. the given Prolog program could successfully be proven:

(0) Obligation:

Clauses:

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

Queries:

sum(a,a,g).

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(2) Obligation:

Triples:

sum1(T28, s(s(T29)), s(s(T27))) :- sum1(T28, T29, T27).
sum1(T54, s(s(T55)), s(s(T53))) :- sum1(T54, T55, T53).

Clauses:

sumc1(T5, 0, T5).
sumc1(T18, s(0), s(T18)).
sumc1(T28, s(s(T29)), s(s(T27))) :- sumc1(T28, T29, T27).
sumc1(T44, s(0), s(T44)).
sumc1(T54, s(s(T55)), s(s(T53))) :- sumc1(T54, T55, T53).

Afs:

sum1(x1, x2, x3)  =  sum1(x3)

(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:
sum1_in: (f,f,b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

SUM1_IN_AAG(T28, s(s(T29)), s(s(T27))) → U1_AAG(T28, T29, T27, sum1_in_aag(T28, T29, T27))
SUM1_IN_AAG(T28, s(s(T29)), s(s(T27))) → SUM1_IN_AAG(T28, T29, T27)
SUM1_IN_AAG(T54, s(s(T55)), s(s(T53))) → U2_AAG(T54, T55, T53, sum1_in_aag(T54, T55, T53))

R is empty.
The argument filtering Pi contains the following mapping:
sum1_in_aag(x1, x2, x3)  =  sum1_in_aag(x3)
s(x1)  =  s(x1)
SUM1_IN_AAG(x1, x2, x3)  =  SUM1_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4)  =  U1_AAG(x3, x4)
U2_AAG(x1, x2, x3, x4)  =  U2_AAG(x3, 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:

SUM1_IN_AAG(T28, s(s(T29)), s(s(T27))) → U1_AAG(T28, T29, T27, sum1_in_aag(T28, T29, T27))
SUM1_IN_AAG(T28, s(s(T29)), s(s(T27))) → SUM1_IN_AAG(T28, T29, T27)
SUM1_IN_AAG(T54, s(s(T55)), s(s(T53))) → U2_AAG(T54, T55, T53, sum1_in_aag(T54, T55, T53))

R is empty.
The argument filtering Pi contains the following mapping:
sum1_in_aag(x1, x2, x3)  =  sum1_in_aag(x3)
s(x1)  =  s(x1)
SUM1_IN_AAG(x1, x2, x3)  =  SUM1_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4)  =  U1_AAG(x3, x4)
U2_AAG(x1, x2, x3, x4)  =  U2_AAG(x3, 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 2 less nodes.

(6) Obligation:

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

SUM1_IN_AAG(T28, s(s(T29)), s(s(T27))) → SUM1_IN_AAG(T28, T29, T27)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
SUM1_IN_AAG(x1, x2, x3)  =  SUM1_IN_AAG(x3)

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:

SUM1_IN_AAG(s(s(T27))) → SUM1_IN_AAG(T27)

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:

• SUM1_IN_AAG(s(s(T27))) → SUM1_IN_AAG(T27)
  The graph contains the following edges 1 > 1

(10) YES