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



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

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

Queries:

sum(a,a,g).

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
sum_in: (f,f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

sum_in_aag(X, s(Y), s(Z)) → U1_aag(X, Y, Z, sum_in_aag(X, Y, Z))
sum_in_aag(X, 0, X) → sum_out_aag(X, 0, X)
U1_aag(X, Y, Z, sum_out_aag(X, Y, Z)) → sum_out_aag(X, s(Y), s(Z))

The argument filtering Pi contains the following mapping:
sum_in_aag(x1, x2, x3)  =  sum_in_aag(x3)
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4)  =  U1_aag(x4)
sum_out_aag(x1, x2, x3)  =  sum_out_aag(x1, x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

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

sum_in_aag(X, s(Y), s(Z)) → U1_aag(X, Y, Z, sum_in_aag(X, Y, Z))
sum_in_aag(X, 0, X) → sum_out_aag(X, 0, X)
U1_aag(X, Y, Z, sum_out_aag(X, Y, Z)) → sum_out_aag(X, s(Y), s(Z))

The argument filtering Pi contains the following mapping:
sum_in_aag(x1, x2, x3)  =  sum_in_aag(x3)
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4)  =  U1_aag(x4)
sum_out_aag(x1, x2, x3)  =  sum_out_aag(x1, x2)


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

SUM_IN_AAG(X, s(Y), s(Z)) → U1_AAG(X, Y, Z, sum_in_aag(X, Y, Z))
SUM_IN_AAG(X, s(Y), s(Z)) → SUM_IN_AAG(X, Y, Z)

The TRS R consists of the following rules:

sum_in_aag(X, s(Y), s(Z)) → U1_aag(X, Y, Z, sum_in_aag(X, Y, Z))
sum_in_aag(X, 0, X) → sum_out_aag(X, 0, X)
U1_aag(X, Y, Z, sum_out_aag(X, Y, Z)) → sum_out_aag(X, s(Y), s(Z))

The argument filtering Pi contains the following mapping:
sum_in_aag(x1, x2, x3)  =  sum_in_aag(x3)
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4)  =  U1_aag(x4)
sum_out_aag(x1, x2, x3)  =  sum_out_aag(x1, x2)
SUM_IN_AAG(x1, x2, x3)  =  SUM_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4)  =  U1_AAG(x4)

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

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof

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

SUM_IN_AAG(X, s(Y), s(Z)) → U1_AAG(X, Y, Z, sum_in_aag(X, Y, Z))
SUM_IN_AAG(X, s(Y), s(Z)) → SUM_IN_AAG(X, Y, Z)

The TRS R consists of the following rules:

sum_in_aag(X, s(Y), s(Z)) → U1_aag(X, Y, Z, sum_in_aag(X, Y, Z))
sum_in_aag(X, 0, X) → sum_out_aag(X, 0, X)
U1_aag(X, Y, Z, sum_out_aag(X, Y, Z)) → sum_out_aag(X, s(Y), s(Z))

The argument filtering Pi contains the following mapping:
sum_in_aag(x1, x2, x3)  =  sum_in_aag(x3)
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4)  =  U1_aag(x4)
sum_out_aag(x1, x2, x3)  =  sum_out_aag(x1, x2)
SUM_IN_AAG(x1, x2, x3)  =  SUM_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4)  =  U1_AAG(x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 1 less node.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof

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

SUM_IN_AAG(X, s(Y), s(Z)) → SUM_IN_AAG(X, Y, Z)

The TRS R consists of the following rules:

sum_in_aag(X, s(Y), s(Z)) → U1_aag(X, Y, Z, sum_in_aag(X, Y, Z))
sum_in_aag(X, 0, X) → sum_out_aag(X, 0, X)
U1_aag(X, Y, Z, sum_out_aag(X, Y, Z)) → sum_out_aag(X, s(Y), s(Z))

The argument filtering Pi contains the following mapping:
sum_in_aag(x1, x2, x3)  =  sum_in_aag(x3)
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4)  =  U1_aag(x4)
sum_out_aag(x1, x2, x3)  =  sum_out_aag(x1, x2)
SUM_IN_AAG(x1, x2, x3)  =  SUM_IN_AAG(x3)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

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

SUM_IN_AAG(X, s(Y), s(Z)) → SUM_IN_AAG(X, Y, Z)

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

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ QDPSizeChangeProof

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

SUM_IN_AAG(s(Z)) → SUM_IN_AAG(Z)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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: