Left Termination of the query pattern p_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

p(X, X).
p(f(X), g(Y)) :- ','(p(f(X), f(Z)), p(Z, g(Y))).

Queries:

p(g,a).

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

p_in_ga(X, X) → p_out_ga(X, X)
p_in_ga(f(X), g(Y)) → U1_ga(X, Y, p_in_ga(f(X), f(Z)))
U1_ga(X, Y, p_out_ga(f(X), f(Z))) → U2_ga(X, Y, Z, p_in_ga(Z, g(Y)))
U2_ga(X, Y, Z, p_out_ga(Z, g(Y))) → p_out_ga(f(X), g(Y))

The argument filtering Pi contains the following mapping:
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
f(x1)  =  f(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x4)
g(x1)  =  g(x1)

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:

p_in_ga(X, X) → p_out_ga(X, X)
p_in_ga(f(X), g(Y)) → U1_ga(X, Y, p_in_ga(f(X), f(Z)))
U1_ga(X, Y, p_out_ga(f(X), f(Z))) → U2_ga(X, Y, Z, p_in_ga(Z, g(Y)))
U2_ga(X, Y, Z, p_out_ga(Z, g(Y))) → p_out_ga(f(X), g(Y))

The argument filtering Pi contains the following mapping:
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
f(x1)  =  f(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x4)
g(x1)  =  g(x1)


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

P_IN_GA(f(X), g(Y)) → U1_GA(X, Y, p_in_ga(f(X), f(Z)))
P_IN_GA(f(X), g(Y)) → P_IN_GA(f(X), f(Z))
U1_GA(X, Y, p_out_ga(f(X), f(Z))) → U2_GA(X, Y, Z, p_in_ga(Z, g(Y)))
U1_GA(X, Y, p_out_ga(f(X), f(Z))) → P_IN_GA(Z, g(Y))

The TRS R consists of the following rules:

p_in_ga(X, X) → p_out_ga(X, X)
p_in_ga(f(X), g(Y)) → U1_ga(X, Y, p_in_ga(f(X), f(Z)))
U1_ga(X, Y, p_out_ga(f(X), f(Z))) → U2_ga(X, Y, Z, p_in_ga(Z, g(Y)))
U2_ga(X, Y, Z, p_out_ga(Z, g(Y))) → p_out_ga(f(X), g(Y))

The argument filtering Pi contains the following mapping:
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
f(x1)  =  f(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x4)
g(x1)  =  g(x1)
P_IN_GA(x1, x2)  =  P_IN_GA(x1)
U2_GA(x1, x2, x3, x4)  =  U2_GA(x4)
U1_GA(x1, x2, x3)  =  U1_GA(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:

P_IN_GA(f(X), g(Y)) → U1_GA(X, Y, p_in_ga(f(X), f(Z)))
P_IN_GA(f(X), g(Y)) → P_IN_GA(f(X), f(Z))
U1_GA(X, Y, p_out_ga(f(X), f(Z))) → U2_GA(X, Y, Z, p_in_ga(Z, g(Y)))
U1_GA(X, Y, p_out_ga(f(X), f(Z))) → P_IN_GA(Z, g(Y))

The TRS R consists of the following rules:

p_in_ga(X, X) → p_out_ga(X, X)
p_in_ga(f(X), g(Y)) → U1_ga(X, Y, p_in_ga(f(X), f(Z)))
U1_ga(X, Y, p_out_ga(f(X), f(Z))) → U2_ga(X, Y, Z, p_in_ga(Z, g(Y)))
U2_ga(X, Y, Z, p_out_ga(Z, g(Y))) → p_out_ga(f(X), g(Y))

The argument filtering Pi contains the following mapping:
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
f(x1)  =  f(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x4)
g(x1)  =  g(x1)
P_IN_GA(x1, x2)  =  P_IN_GA(x1)
U2_GA(x1, x2, x3, x4)  =  U2_GA(x4)
U1_GA(x1, x2, x3)  =  U1_GA(x3)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 2 less nodes.

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

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

U1_GA(X, Y, p_out_ga(f(X), f(Z))) → P_IN_GA(Z, g(Y))
P_IN_GA(f(X), g(Y)) → U1_GA(X, Y, p_in_ga(f(X), f(Z)))

The TRS R consists of the following rules:

p_in_ga(X, X) → p_out_ga(X, X)
p_in_ga(f(X), g(Y)) → U1_ga(X, Y, p_in_ga(f(X), f(Z)))
U1_ga(X, Y, p_out_ga(f(X), f(Z))) → U2_ga(X, Y, Z, p_in_ga(Z, g(Y)))
U2_ga(X, Y, Z, p_out_ga(Z, g(Y))) → p_out_ga(f(X), g(Y))

The argument filtering Pi contains the following mapping:
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
f(x1)  =  f(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x4)
g(x1)  =  g(x1)
P_IN_GA(x1, x2)  =  P_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(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:

U1_GA(X, Y, p_out_ga(f(X), f(Z))) → P_IN_GA(Z, g(Y))
P_IN_GA(f(X), g(Y)) → U1_GA(X, Y, p_in_ga(f(X), f(Z)))

The TRS R consists of the following rules:

p_in_ga(X, X) → p_out_ga(X, X)

The argument filtering Pi contains the following mapping:
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
f(x1)  =  f(x1)
g(x1)  =  g(x1)
P_IN_GA(x1, x2)  =  P_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(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
                      ↳ RuleRemovalProof

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

P_IN_GA(f(X)) → U1_GA(p_in_ga(f(X)))
U1_GA(p_out_ga(f(Z))) → P_IN_GA(Z)

The TRS R consists of the following rules:

p_in_ga(X) → p_out_ga(X)

The set Q consists of the following terms:

p_in_ga(x0)

We have to consider all (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

P_IN_GA(f(X)) → U1_GA(p_in_ga(f(X)))
U1_GA(p_out_ga(f(Z))) → P_IN_GA(Z)


Used ordering: POLO with Polynomial interpretation [25]:

POL(P_IN_GA(x1)) = 2 + 2·x1   
POL(U1_GA(x1)) = 1 + 2·x1   
POL(f(x1)) = 1 + 2·x1   
POL(p_in_ga(x1)) = x1   
POL(p_out_ga(x1)) = x1   



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ RuleRemovalProof
                        ↳ QDP
                          ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

p_in_ga(X) → p_out_ga(X)

The set Q consists of the following terms:

p_in_ga(x0)

We have to consider all (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.