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

0 Prolog
  ↳1 PrologToPiTRSProof (⇐)
    ↳2 PiTRS
    ↳3 DependencyPairsProof (⇔)
    ↳4 PiDP
    ↳5 DependencyGraphProof (⇔)
    ↳6 PiDP
    ↳7 UsableRulesProof (⇔)
    ↳8 PiDP
    ↳9 PiDPToQDPProof (⇔)
    ↳10 QDP
    ↳11 DependencyGraphProof (⇔)
    ↳12 TRUE
  ↳13 PrologToPiTRSProof (⇐)
    ↳14 PiTRS
    ↳15 DependencyPairsProof (⇔)
    ↳16 PiDP
    ↳17 DependencyGraphProof (⇔)
    ↳18 PiDP
    ↳19 UsableRulesProof (⇔)
    ↳20 PiDP
    ↳21 PiDPToQDPProof (⇔)
    ↳22 QDP

(0) Obligation:

Clauses:

f(0, 1, X) :- f(X, X, X).

Queries:

f(a,a,g).

(1) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. 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

(2) Obligation:

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)

(3) 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:

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_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)
U1_GGG(x1, x2)  =  U1_GGG(x2)

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

(4) Obligation:

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_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)
U1_GGG(x1, x2)  =  U1_GGG(x2)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.

(6) Obligation:

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

(7) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(8) Obligation:

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

(9) PiDPToQDPProof (EQUIVALENT transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(10) Obligation:

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.

(11) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(12) TRUE

(13) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. 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(x1, x2)
f_in_ggg(x1, x2, x3)  =  f_in_ggg(x1, x2, x3)
0  =  0
1  =  1
U1_ggg(x1, x2)  =  U1_ggg(x1, x2)
f_out_ggg(x1, x2, x3)  =  f_out_ggg(x1, x2, x3)
f_out_aag(x1, x2, x3)  =  f_out_aag(x1, x2, x3)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(14) Obligation:

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(x1, x2)
f_in_ggg(x1, x2, x3)  =  f_in_ggg(x1, x2, x3)
0  =  0
1  =  1
U1_ggg(x1, x2)  =  U1_ggg(x1, x2)
f_out_ggg(x1, x2, x3)  =  f_out_ggg(x1, x2, x3)
f_out_aag(x1, x2, x3)  =  f_out_aag(x1, x2, x3)

(15) 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:

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(x1, x2)
f_in_ggg(x1, x2, x3)  =  f_in_ggg(x1, x2, x3)
0  =  0
1  =  1
U1_ggg(x1, x2)  =  U1_ggg(x1, x2)
f_out_ggg(x1, x2, x3)  =  f_out_ggg(x1, x2, x3)
f_out_aag(x1, x2, x3)  =  f_out_aag(x1, x2, x3)
F_IN_AAG(x1, x2, x3)  =  F_IN_AAG(x3)
U1_AAG(x1, x2)  =  U1_AAG(x1, x2)
F_IN_GGG(x1, x2, x3)  =  F_IN_GGG(x1, x2, x3)
U1_GGG(x1, x2)  =  U1_GGG(x1, x2)

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

(16) Obligation:

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(x1, x2)
f_in_ggg(x1, x2, x3)  =  f_in_ggg(x1, x2, x3)
0  =  0
1  =  1
U1_ggg(x1, x2)  =  U1_ggg(x1, x2)
f_out_ggg(x1, x2, x3)  =  f_out_ggg(x1, x2, x3)
f_out_aag(x1, x2, x3)  =  f_out_aag(x1, x2, x3)
F_IN_AAG(x1, x2, x3)  =  F_IN_AAG(x3)
U1_AAG(x1, x2)  =  U1_AAG(x1, x2)
F_IN_GGG(x1, x2, x3)  =  F_IN_GGG(x1, x2, x3)
U1_GGG(x1, x2)  =  U1_GGG(x1, x2)

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

(17) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.

(18) Obligation:

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(x1, x2)
f_in_ggg(x1, x2, x3)  =  f_in_ggg(x1, x2, x3)
0  =  0
1  =  1
U1_ggg(x1, x2)  =  U1_ggg(x1, x2)
f_out_ggg(x1, x2, x3)  =  f_out_ggg(x1, x2, x3)
f_out_aag(x1, x2, x3)  =  f_out_aag(x1, x2, x3)
F_IN_GGG(x1, x2, x3)  =  F_IN_GGG(x1, x2, x3)

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

(19) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(20) Obligation:

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

(21) PiDPToQDPProof (EQUIVALENT transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(22) Obligation:

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.