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



↳ PROLOG
  ↳ PrologToPiTRSProof

f3(00, 10, X) :- f3(X, X, X).


With regard to the inferred argument filtering the predicates were used in the following modes:
f3: (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_3_in_aag3(0_0, 1_0, X) -> if_f_3_in_1_aag2(X, f_3_in_ggg3(X, X, X))
f_3_in_ggg3(0_0, 1_0, X) -> if_f_3_in_1_ggg2(X, f_3_in_ggg3(X, X, X))
if_f_3_in_1_ggg2(X, f_3_out_ggg3(X, X, X)) -> f_3_out_ggg3(0_0, 1_0, X)
if_f_3_in_1_aag2(X, f_3_out_ggg3(X, X, X)) -> f_3_out_aag3(0_0, 1_0, X)

The argument filtering Pi contains the following mapping:
f_3_in_aag3(x1, x2, x3)  =  f_3_in_aag1(x3)
0_0  =  0_0
1_0  =  1_0
if_f_3_in_1_aag2(x1, x2)  =  if_f_3_in_1_aag1(x2)
f_3_in_ggg3(x1, x2, x3)  =  f_3_in_ggg3(x1, x2, x3)
if_f_3_in_1_ggg2(x1, x2)  =  if_f_3_in_1_ggg1(x2)
f_3_out_ggg3(x1, x2, x3)  =  f_3_out_ggg
f_3_out_aag3(x1, x2, x3)  =  f_3_out_aag2(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_3_in_aag3(0_0, 1_0, X) -> if_f_3_in_1_aag2(X, f_3_in_ggg3(X, X, X))
f_3_in_ggg3(0_0, 1_0, X) -> if_f_3_in_1_ggg2(X, f_3_in_ggg3(X, X, X))
if_f_3_in_1_ggg2(X, f_3_out_ggg3(X, X, X)) -> f_3_out_ggg3(0_0, 1_0, X)
if_f_3_in_1_aag2(X, f_3_out_ggg3(X, X, X)) -> f_3_out_aag3(0_0, 1_0, X)

The argument filtering Pi contains the following mapping:
f_3_in_aag3(x1, x2, x3)  =  f_3_in_aag1(x3)
0_0  =  0_0
1_0  =  1_0
if_f_3_in_1_aag2(x1, x2)  =  if_f_3_in_1_aag1(x2)
f_3_in_ggg3(x1, x2, x3)  =  f_3_in_ggg3(x1, x2, x3)
if_f_3_in_1_ggg2(x1, x2)  =  if_f_3_in_1_ggg1(x2)
f_3_out_ggg3(x1, x2, x3)  =  f_3_out_ggg
f_3_out_aag3(x1, x2, x3)  =  f_3_out_aag2(x1, x2)


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

F_3_IN_AAG3(0_0, 1_0, X) -> IF_F_3_IN_1_AAG2(X, f_3_in_ggg3(X, X, X))
F_3_IN_AAG3(0_0, 1_0, X) -> F_3_IN_GGG3(X, X, X)
F_3_IN_GGG3(0_0, 1_0, X) -> IF_F_3_IN_1_GGG2(X, f_3_in_ggg3(X, X, X))
F_3_IN_GGG3(0_0, 1_0, X) -> F_3_IN_GGG3(X, X, X)

The TRS R consists of the following rules:

f_3_in_aag3(0_0, 1_0, X) -> if_f_3_in_1_aag2(X, f_3_in_ggg3(X, X, X))
f_3_in_ggg3(0_0, 1_0, X) -> if_f_3_in_1_ggg2(X, f_3_in_ggg3(X, X, X))
if_f_3_in_1_ggg2(X, f_3_out_ggg3(X, X, X)) -> f_3_out_ggg3(0_0, 1_0, X)
if_f_3_in_1_aag2(X, f_3_out_ggg3(X, X, X)) -> f_3_out_aag3(0_0, 1_0, X)

The argument filtering Pi contains the following mapping:
f_3_in_aag3(x1, x2, x3)  =  f_3_in_aag1(x3)
0_0  =  0_0
1_0  =  1_0
if_f_3_in_1_aag2(x1, x2)  =  if_f_3_in_1_aag1(x2)
f_3_in_ggg3(x1, x2, x3)  =  f_3_in_ggg3(x1, x2, x3)
if_f_3_in_1_ggg2(x1, x2)  =  if_f_3_in_1_ggg1(x2)
f_3_out_ggg3(x1, x2, x3)  =  f_3_out_ggg
f_3_out_aag3(x1, x2, x3)  =  f_3_out_aag2(x1, x2)
F_3_IN_AAG3(x1, x2, x3)  =  F_3_IN_AAG1(x3)
IF_F_3_IN_1_AAG2(x1, x2)  =  IF_F_3_IN_1_AAG1(x2)
IF_F_3_IN_1_GGG2(x1, x2)  =  IF_F_3_IN_1_GGG1(x2)
F_3_IN_GGG3(x1, x2, x3)  =  F_3_IN_GGG3(x1, x2, x3)

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_3_IN_AAG3(0_0, 1_0, X) -> IF_F_3_IN_1_AAG2(X, f_3_in_ggg3(X, X, X))
F_3_IN_AAG3(0_0, 1_0, X) -> F_3_IN_GGG3(X, X, X)
F_3_IN_GGG3(0_0, 1_0, X) -> IF_F_3_IN_1_GGG2(X, f_3_in_ggg3(X, X, X))
F_3_IN_GGG3(0_0, 1_0, X) -> F_3_IN_GGG3(X, X, X)

The TRS R consists of the following rules:

f_3_in_aag3(0_0, 1_0, X) -> if_f_3_in_1_aag2(X, f_3_in_ggg3(X, X, X))
f_3_in_ggg3(0_0, 1_0, X) -> if_f_3_in_1_ggg2(X, f_3_in_ggg3(X, X, X))
if_f_3_in_1_ggg2(X, f_3_out_ggg3(X, X, X)) -> f_3_out_ggg3(0_0, 1_0, X)
if_f_3_in_1_aag2(X, f_3_out_ggg3(X, X, X)) -> f_3_out_aag3(0_0, 1_0, X)

The argument filtering Pi contains the following mapping:
f_3_in_aag3(x1, x2, x3)  =  f_3_in_aag1(x3)
0_0  =  0_0
1_0  =  1_0
if_f_3_in_1_aag2(x1, x2)  =  if_f_3_in_1_aag1(x2)
f_3_in_ggg3(x1, x2, x3)  =  f_3_in_ggg3(x1, x2, x3)
if_f_3_in_1_ggg2(x1, x2)  =  if_f_3_in_1_ggg1(x2)
f_3_out_ggg3(x1, x2, x3)  =  f_3_out_ggg
f_3_out_aag3(x1, x2, x3)  =  f_3_out_aag2(x1, x2)
F_3_IN_AAG3(x1, x2, x3)  =  F_3_IN_AAG1(x3)
IF_F_3_IN_1_AAG2(x1, x2)  =  IF_F_3_IN_1_AAG1(x2)
IF_F_3_IN_1_GGG2(x1, x2)  =  IF_F_3_IN_1_GGG1(x2)
F_3_IN_GGG3(x1, x2, x3)  =  F_3_IN_GGG3(x1, x2, x3)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph 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_3_IN_GGG3(0_0, 1_0, X) -> F_3_IN_GGG3(X, X, X)

The TRS R consists of the following rules:

f_3_in_aag3(0_0, 1_0, X) -> if_f_3_in_1_aag2(X, f_3_in_ggg3(X, X, X))
f_3_in_ggg3(0_0, 1_0, X) -> if_f_3_in_1_ggg2(X, f_3_in_ggg3(X, X, X))
if_f_3_in_1_ggg2(X, f_3_out_ggg3(X, X, X)) -> f_3_out_ggg3(0_0, 1_0, X)
if_f_3_in_1_aag2(X, f_3_out_ggg3(X, X, X)) -> f_3_out_aag3(0_0, 1_0, X)

The argument filtering Pi contains the following mapping:
f_3_in_aag3(x1, x2, x3)  =  f_3_in_aag1(x3)
0_0  =  0_0
1_0  =  1_0
if_f_3_in_1_aag2(x1, x2)  =  if_f_3_in_1_aag1(x2)
f_3_in_ggg3(x1, x2, x3)  =  f_3_in_ggg3(x1, x2, x3)
if_f_3_in_1_ggg2(x1, x2)  =  if_f_3_in_1_ggg1(x2)
f_3_out_ggg3(x1, x2, x3)  =  f_3_out_ggg
f_3_out_aag3(x1, x2, x3)  =  f_3_out_aag2(x1, x2)
F_3_IN_GGG3(x1, x2, x3)  =  F_3_IN_GGG3(x1, x2, x3)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting 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_3_IN_GGG3(0_0, 1_0, X) -> F_3_IN_GGG3(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 into ordinary QDP problem 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_3_IN_GGG3(0_0, 1_0, X) -> F_3_IN_GGG3(X, X, X)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {F_3_IN_GGG3}.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.