Left Termination of the query pattern f_in_3(a, a, g) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

f(0, 1, X) :- f(X, X, X).

Queries:

f(a,a,g).

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

f_in_aag(0, 1, X) → U1_aag(X, f_in_ggg(X, X, X))
f_in_ggg(0, 1, X) → U1_ggg(X, f_in_ggg(X, X, X))
U1_ggg(X, f_out_ggg(X, X, X)) → f_out_ggg(0, 1, X)
U1_aag(X, f_out_ggg(X, X, X)) → f_out_aag(0, 1, X)

The argument filtering Pi contains the following mapping:
f_in_aag(x1, x2, x3)  =  f_in_aag(x3)
U1_aag(x1, x2)  =  U1_aag(x2)
f_in_ggg(x1, x2, x3)  =  f_in_ggg(x1, x2, x3)
0  =  0
1  =  1
U1_ggg(x1, x2)  =  U1_ggg(x2)
f_out_ggg(x1, x2, x3)  =  f_out_ggg
f_out_aag(x1, x2, x3)  =  f_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:

f_in_aag(0, 1, X) → U1_aag(X, f_in_ggg(X, X, X))
f_in_ggg(0, 1, X) → U1_ggg(X, f_in_ggg(X, X, X))
U1_ggg(X, f_out_ggg(X, X, X)) → f_out_ggg(0, 1, X)
U1_aag(X, f_out_ggg(X, X, X)) → f_out_aag(0, 1, X)

The argument filtering Pi contains the following mapping:
f_in_aag(x1, x2, x3)  =  f_in_aag(x3)
U1_aag(x1, x2)  =  U1_aag(x2)
f_in_ggg(x1, x2, x3)  =  f_in_ggg(x1, x2, x3)
0  =  0
1  =  1
U1_ggg(x1, x2)  =  U1_ggg(x2)
f_out_ggg(x1, x2, x3)  =  f_out_ggg
f_out_aag(x1, x2, x3)  =  f_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:

F_IN_AAG(0, 1, X) → U1_AAG(X, f_in_ggg(X, X, X))
F_IN_AAG(0, 1, X) → F_IN_GGG(X, X, X)
F_IN_GGG(0, 1, X) → U1_GGG(X, f_in_ggg(X, X, X))
F_IN_GGG(0, 1, X) → F_IN_GGG(X, X, X)

The TRS R consists of the following rules:

f_in_aag(0, 1, X) → U1_aag(X, f_in_ggg(X, X, X))
f_in_ggg(0, 1, X) → U1_ggg(X, f_in_ggg(X, X, X))
U1_ggg(X, f_out_ggg(X, X, X)) → f_out_ggg(0, 1, X)
U1_aag(X, f_out_ggg(X, X, X)) → f_out_aag(0, 1, X)

The argument filtering Pi contains the following mapping:
f_in_aag(x1, x2, x3)  =  f_in_aag(x3)
U1_aag(x1, x2)  =  U1_aag(x2)
f_in_ggg(x1, x2, x3)  =  f_in_ggg(x1, x2, x3)
0  =  0
1  =  1
U1_ggg(x1, x2)  =  U1_ggg(x2)
f_out_ggg(x1, x2, x3)  =  f_out_ggg
f_out_aag(x1, x2, x3)  =  f_out_aag(x1, x2)
F_IN_GGG(x1, x2, x3)  =  F_IN_GGG(x1, x2, x3)
U1_AAG(x1, x2)  =  U1_AAG(x2)
F_IN_AAG(x1, x2, x3)  =  F_IN_AAG(x3)
U1_GGG(x1, x2)  =  U1_GGG(x2)

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:

F_IN_AAG(0, 1, X) → U1_AAG(X, f_in_ggg(X, X, X))
F_IN_AAG(0, 1, X) → F_IN_GGG(X, X, X)
F_IN_GGG(0, 1, X) → U1_GGG(X, f_in_ggg(X, X, X))
F_IN_GGG(0, 1, X) → F_IN_GGG(X, X, X)

The TRS R consists of the following rules:

f_in_aag(0, 1, X) → U1_aag(X, f_in_ggg(X, X, X))
f_in_ggg(0, 1, X) → U1_ggg(X, f_in_ggg(X, X, X))
U1_ggg(X, f_out_ggg(X, X, X)) → f_out_ggg(0, 1, X)
U1_aag(X, f_out_ggg(X, X, X)) → f_out_aag(0, 1, X)

The argument filtering Pi contains the following mapping:
f_in_aag(x1, x2, x3)  =  f_in_aag(x3)
U1_aag(x1, x2)  =  U1_aag(x2)
f_in_ggg(x1, x2, x3)  =  f_in_ggg(x1, x2, x3)
0  =  0
1  =  1
U1_ggg(x1, x2)  =  U1_ggg(x2)
f_out_ggg(x1, x2, x3)  =  f_out_ggg
f_out_aag(x1, x2, x3)  =  f_out_aag(x1, x2)
F_IN_GGG(x1, x2, x3)  =  F_IN_GGG(x1, x2, x3)
U1_AAG(x1, x2)  =  U1_AAG(x2)
F_IN_AAG(x1, x2, x3)  =  F_IN_AAG(x3)
U1_GGG(x1, x2)  =  U1_GGG(x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 3 less nodes.

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

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

F_IN_GGG(0, 1, X) → F_IN_GGG(X, X, X)

The TRS R consists of the following rules:

f_in_aag(0, 1, X) → U1_aag(X, f_in_ggg(X, X, X))
f_in_ggg(0, 1, X) → U1_ggg(X, f_in_ggg(X, X, X))
U1_ggg(X, f_out_ggg(X, X, X)) → f_out_ggg(0, 1, X)
U1_aag(X, f_out_ggg(X, X, X)) → f_out_aag(0, 1, X)

The argument filtering Pi contains the following mapping:
f_in_aag(x1, x2, x3)  =  f_in_aag(x3)
U1_aag(x1, x2)  =  U1_aag(x2)
f_in_ggg(x1, x2, x3)  =  f_in_ggg(x1, x2, x3)
0  =  0
1  =  1
U1_ggg(x1, x2)  =  U1_ggg(x2)
f_out_ggg(x1, x2, x3)  =  f_out_ggg
f_out_aag(x1, x2, x3)  =  f_out_aag(x1, x2)
F_IN_GGG(x1, x2, x3)  =  F_IN_GGG(x1, x2, 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:

F_IN_GGG(0, 1, X) → F_IN_GGG(X, X, X)

R is empty.
Pi is empty.
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
                      ↳ DependencyGraphProof

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

F_IN_GGG(0, 1, X) → F_IN_GGG(X, X, X)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.