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



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

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

Queries:

plus(g,a,a).

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

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, Z) → U1_gaa(X, Y, Z, plus_in_gaa(X, s(Y), Z))
U1_gaa(X, Y, Z, plus_out_gaa(X, s(Y), Z)) → plus_out_gaa(s(X), Y, Z)

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_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



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

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

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, Z) → U1_gaa(X, Y, Z, plus_in_gaa(X, s(Y), Z))
U1_gaa(X, Y, Z, plus_out_gaa(X, s(Y), Z)) → plus_out_gaa(s(X), Y, Z)

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


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

PLUS_IN_GAA(s(X), Y, Z) → U1_GAA(X, Y, Z, plus_in_gaa(X, s(Y), Z))
PLUS_IN_GAA(s(X), Y, Z) → PLUS_IN_GAA(X, s(Y), Z)

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, Z) → U1_gaa(X, Y, Z, plus_in_gaa(X, s(Y), Z))
U1_gaa(X, Y, Z, plus_out_gaa(X, s(Y), Z)) → plus_out_gaa(s(X), Y, Z)

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa
s(x1)  =  s(x1)
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x4)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x4)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)

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:

PLUS_IN_GAA(s(X), Y, Z) → U1_GAA(X, Y, Z, plus_in_gaa(X, s(Y), Z))
PLUS_IN_GAA(s(X), Y, Z) → PLUS_IN_GAA(X, s(Y), Z)

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, Z) → U1_gaa(X, Y, Z, plus_in_gaa(X, s(Y), Z))
U1_gaa(X, Y, Z, plus_out_gaa(X, s(Y), Z)) → plus_out_gaa(s(X), Y, Z)

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa
s(x1)  =  s(x1)
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x4)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x4)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)

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:

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

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, Z) → U1_gaa(X, Y, Z, plus_in_gaa(X, s(Y), Z))
U1_gaa(X, Y, Z, plus_out_gaa(X, s(Y), Z)) → plus_out_gaa(s(X), Y, Z)

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

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:

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

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

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:

PLUS_IN_GAA(s(X)) → PLUS_IN_GAA(X)

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: