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

0 Prolog
↳1 PrologToPrologProblemTransformerProof (⇐)
↳2 Prolog
↳3 PrologToPiTRSProof (⇐)
↳4 PiTRS
↳5 DependencyPairsProof (⇔)
↳6 PiDP
↳7 DependencyGraphProof (⇔)
↳8 PiDP
↳9 UsableRulesProof (⇔)
↳10 PiDP
↳11 PiDPToQDPProof (⇐)
↳12 QDP
↳13 QDPSizeChangeProof (⇔)
↳14 YES

(0) Obligation:

Clauses:

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

Queries:

sum(a,a,g).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

(2) Obligation:

Clauses:

sum1(T5, 0, T5).
sum1(T18, s(0), s(T18)).
sum1(T28, s(s(T29)), s(s(T27))) :- sum1(T28, T29, T27).
sum1(T44, s(0), s(T44)).
sum1(T54, s(s(T55)), s(s(T53))) :- sum1(T54, T55, T53).

Queries:

sum1(a,a,g).

(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:
sum1_in: (f,f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

sum1_in_aag(T5, 0, T5) → sum1_out_aag(T5, 0, T5)
sum1_in_aag(T18, s(0), s(T18)) → sum1_out_aag(T18, s(0), s(T18))
sum1_in_aag(T28, s(s(T29)), s(s(T27))) → U1_aag(T28, T29, 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))
U2_aag(T54, T55, T53, sum1_out_aag(T54, T55, T53)) → sum1_out_aag(T54, s(s(T55)), s(s(T53)))
U1_aag(T28, T29, T27, sum1_out_aag(T28, T29, T27)) → sum1_out_aag(T28, s(s(T29)), s(s(T27)))

The argument filtering Pi contains the following mapping:
sum1_in_aag(x1, x2, x3)  =  sum1_in_aag(x3)
sum1_out_aag(x1, x2, x3)  =  sum1_out_aag(x1, x2)
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4)  =  U1_aag(x4)
U2_aag(x1, x2, x3, x4)  =  U2_aag(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:

sum1_in_aag(T5, 0, T5) → sum1_out_aag(T5, 0, T5)
sum1_in_aag(T18, s(0), s(T18)) → sum1_out_aag(T18, s(0), s(T18))
sum1_in_aag(T28, s(s(T29)), s(s(T27))) → U1_aag(T28, T29, 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))
U2_aag(T54, T55, T53, sum1_out_aag(T54, T55, T53)) → sum1_out_aag(T54, s(s(T55)), s(s(T53)))
U1_aag(T28, T29, T27, sum1_out_aag(T28, T29, T27)) → sum1_out_aag(T28, s(s(T29)), s(s(T27)))

The argument filtering Pi contains the following mapping:
sum1_in_aag(x1, x2, x3)  =  sum1_in_aag(x3)
sum1_out_aag(x1, x2, x3)  =  sum1_out_aag(x1, x2)
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4)  =  U1_aag(x4)
U2_aag(x1, x2, x3, x4)  =  U2_aag(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:

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))

The TRS R consists of the following rules:

sum1_in_aag(T5, 0, T5) → sum1_out_aag(T5, 0, T5)
sum1_in_aag(T18, s(0), s(T18)) → sum1_out_aag(T18, s(0), s(T18))
sum1_in_aag(T28, s(s(T29)), s(s(T27))) → U1_aag(T28, T29, 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))
U2_aag(T54, T55, T53, sum1_out_aag(T54, T55, T53)) → sum1_out_aag(T54, s(s(T55)), s(s(T53)))
U1_aag(T28, T29, T27, sum1_out_aag(T28, T29, T27)) → sum1_out_aag(T28, s(s(T29)), s(s(T27)))

The argument filtering Pi contains the following mapping:
sum1_in_aag(x1, x2, x3)  =  sum1_in_aag(x3)
sum1_out_aag(x1, x2, x3)  =  sum1_out_aag(x1, x2)
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4)  =  U1_aag(x4)
U2_aag(x1, x2, x3, x4)  =  U2_aag(x4)
SUM1_IN_AAG(x1, x2, x3)  =  SUM1_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4)  =  U1_AAG(x4)
U2_AAG(x1, x2, x3, x4)  =  U2_AAG(x4)

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

(6) 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))

The TRS R consists of the following rules:

sum1_in_aag(T5, 0, T5) → sum1_out_aag(T5, 0, T5)
sum1_in_aag(T18, s(0), s(T18)) → sum1_out_aag(T18, s(0), s(T18))
sum1_in_aag(T28, s(s(T29)), s(s(T27))) → U1_aag(T28, T29, 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))
U2_aag(T54, T55, T53, sum1_out_aag(T54, T55, T53)) → sum1_out_aag(T54, s(s(T55)), s(s(T53)))
U1_aag(T28, T29, T27, sum1_out_aag(T28, T29, T27)) → sum1_out_aag(T28, s(s(T29)), s(s(T27)))

The argument filtering Pi contains the following mapping:
sum1_in_aag(x1, x2, x3)  =  sum1_in_aag(x3)
sum1_out_aag(x1, x2, x3)  =  sum1_out_aag(x1, x2)
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4)  =  U1_aag(x4)
U2_aag(x1, x2, x3, x4)  =  U2_aag(x4)
SUM1_IN_AAG(x1, x2, x3)  =  SUM1_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4)  =  U1_AAG(x4)
U2_AAG(x1, x2, x3, x4)  =  U2_AAG(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 2 less nodes.

(8) 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)

The TRS R consists of the following rules:

sum1_in_aag(T5, 0, T5) → sum1_out_aag(T5, 0, T5)
sum1_in_aag(T18, s(0), s(T18)) → sum1_out_aag(T18, s(0), s(T18))
sum1_in_aag(T28, s(s(T29)), s(s(T27))) → U1_aag(T28, T29, 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))
U2_aag(T54, T55, T53, sum1_out_aag(T54, T55, T53)) → sum1_out_aag(T54, s(s(T55)), s(s(T53)))
U1_aag(T28, T29, T27, sum1_out_aag(T28, T29, T27)) → sum1_out_aag(T28, s(s(T29)), s(s(T27)))

The argument filtering Pi contains the following mapping:
sum1_in_aag(x1, x2, x3)  =  sum1_in_aag(x3)
sum1_out_aag(x1, x2, x3)  =  sum1_out_aag(x1, x2)
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4)  =  U1_aag(x4)
U2_aag(x1, x2, x3, x4)  =  U2_aag(x4)
SUM1_IN_AAG(x1, x2, x3)  =  SUM1_IN_AAG(x3)

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:

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

(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:

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.

(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:

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

(14) YES