Left Termination proof of /tmp/tmpE0jIJD/paper1.pl

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) (b,b)
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_gg(Z, g(Y)))
p_in_gg(X, X) → p_out_gg(X, X)
p_in_gg(f(X), g(Y)) → U1_gg(X, Y, p_in_ga(f(X), f(Z)))
U1_gg(X, Y, p_out_ga(f(X), f(Z))) → U2_gg(X, Y, Z, p_in_gg(Z, g(Y)))
U2_gg(X, Y, Z, p_out_gg(Z, g(Y))) → p_out_gg(f(X), g(Y))
U2_ga(X, Y, Z, p_out_gg(Z, g(Y))) → p_out_ga(f(X), g(Y))

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_gg(Z, g(Y)))
p_in_gg(X, X) → p_out_gg(X, X)
p_in_gg(f(X), g(Y)) → U1_gg(X, Y, p_in_ga(f(X), f(Z)))
U1_gg(X, Y, p_out_ga(f(X), f(Z))) → U2_gg(X, Y, Z, p_in_gg(Z, g(Y)))
U2_gg(X, Y, Z, p_out_gg(Z, g(Y))) → p_out_gg(f(X), g(Y))
U2_ga(X, Y, Z, p_out_gg(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)
p_in_gg(x1, x2) = p_in_gg(x1, x2)
g(x1) = g
p_out_gg(x1, x2) = p_out_gg
U1_gg(x1, x2, x3) = U1_gg(x3)
U2_gg(x1, x2, x3, x4) = U2_gg(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:

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_gg(Z, g(Y)))
U1_GA(X, Y, p_out_ga(f(X), f(Z))) → P_IN_GG(Z, g(Y))
P_IN_GG(f(X), g(Y)) → U1_GG(X, Y, p_in_ga(f(X), f(Z)))
P_IN_GG(f(X), g(Y)) → P_IN_GA(f(X), f(Z))
U1_GG(X, Y, p_out_ga(f(X), f(Z))) → U2_GG(X, Y, Z, p_in_gg(Z, g(Y)))
U1_GG(X, Y, p_out_ga(f(X), f(Z))) → P_IN_GG(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_gg(Z, g(Y)))
p_in_gg(X, X) → p_out_gg(X, X)
p_in_gg(f(X), g(Y)) → U1_gg(X, Y, p_in_ga(f(X), f(Z)))
U1_gg(X, Y, p_out_ga(f(X), f(Z))) → U2_gg(X, Y, Z, p_in_gg(Z, g(Y)))
U2_gg(X, Y, Z, p_out_gg(Z, g(Y))) → p_out_gg(f(X), g(Y))
U2_ga(X, Y, Z, p_out_gg(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)
p_in_gg(x1, x2) = p_in_gg(x1, x2)
g(x1) = g
p_out_gg(x1, x2) = p_out_gg
U1_gg(x1, x2, x3) = U1_gg(x3)
U2_gg(x1, x2, x3, x4) = U2_gg(x4)
P_IN_GA(x1, x2) = P_IN_GA(x1)
U1_GG(x1, x2, x3) = U1_GG(x3)
P_IN_GG(x1, x2) = P_IN_GG(x1, x2)
U2_GG(x1, x2, x3, x4) = U2_GG(x4)
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_gg(Z, g(Y)))
U1_GA(X, Y, p_out_ga(f(X), f(Z))) → P_IN_GG(Z, g(Y))
P_IN_GG(f(X), g(Y)) → U1_GG(X, Y, p_in_ga(f(X), f(Z)))
P_IN_GG(f(X), g(Y)) → P_IN_GA(f(X), f(Z))
U1_GG(X, Y, p_out_ga(f(X), f(Z))) → U2_GG(X, Y, Z, p_in_gg(Z, g(Y)))
U1_GG(X, Y, p_out_ga(f(X), f(Z))) → P_IN_GG(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_gg(Z, g(Y)))
p_in_gg(X, X) → p_out_gg(X, X)
p_in_gg(f(X), g(Y)) → U1_gg(X, Y, p_in_ga(f(X), f(Z)))
U1_gg(X, Y, p_out_ga(f(X), f(Z))) → U2_gg(X, Y, Z, p_in_gg(Z, g(Y)))
U2_gg(X, Y, Z, p_out_gg(Z, g(Y))) → p_out_gg(f(X), g(Y))
U2_ga(X, Y, Z, p_out_gg(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)
p_in_gg(x1, x2) = p_in_gg(x1, x2)
g(x1) = g
p_out_gg(x1, x2) = p_out_gg
U1_gg(x1, x2, x3) = U1_gg(x3)
U2_gg(x1, x2, x3, x4) = U2_gg(x4)
P_IN_GA(x1, x2) = P_IN_GA(x1)
U1_GG(x1, x2, x3) = U1_GG(x3)
P_IN_GG(x1, x2) = P_IN_GG(x1, x2)
U2_GG(x1, x2, x3, x4) = U2_GG(x4)
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 6 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

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

P_IN_GG(f(X), g(Y)) → U1_GG(X, Y, p_in_ga(f(X), f(Z)))
U1_GG(X, Y, p_out_ga(f(X), f(Z))) → P_IN_GG(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_gg(Z, g(Y)))
p_in_gg(X, X) → p_out_gg(X, X)
p_in_gg(f(X), g(Y)) → U1_gg(X, Y, p_in_ga(f(X), f(Z)))
U1_gg(X, Y, p_out_ga(f(X), f(Z))) → U2_gg(X, Y, Z, p_in_gg(Z, g(Y)))
U2_gg(X, Y, Z, p_out_gg(Z, g(Y))) → p_out_gg(f(X), g(Y))
U2_ga(X, Y, Z, p_out_gg(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)
p_in_gg(x1, x2) = p_in_gg(x1, x2)
g(x1) = g
p_out_gg(x1, x2) = p_out_gg
U1_gg(x1, x2, x3) = U1_gg(x3)
U2_gg(x1, x2, x3, x4) = U2_gg(x4)
U1_GG(x1, x2, x3) = U1_GG(x3)
P_IN_GG(x1, x2) = P_IN_GG(x1, x2)

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:

P_IN_GG(f(X), g(Y)) → U1_GG(X, Y, p_in_ga(f(X), f(Z)))
U1_GG(X, Y, p_out_ga(f(X), f(Z))) → P_IN_GG(Z, g(Y))

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
U1_GG(x1, x2, x3) = U1_GG(x3)
P_IN_GG(x1, x2) = P_IN_GG(x1, x2)

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:

U1_GG(p_out_ga(f(Z))) → P_IN_GG(Z, g)
P_IN_GG(f(X), g) → U1_GG(p_in_ga(f(X)))

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:

U1_GG(p_out_ga(f(Z))) → P_IN_GG(Z, g)
P_IN_GG(f(X), g) → U1_GG(p_in_ga(f(X)))

Strictly oriented rules of the TRS R:

p_in_ga(X) → p_out_ga(X)

Used ordering: POLO with Polynomial interpretation [25]:

POL(P_IN_GG(x₁, x₂)) = 2·x₁ + 2·x₂
POL(U1_GG(x₁)) = 1 + x₁
POL(f(x₁)) = 2 + x₁
POL(g) = 2
POL(p_in_ga(x₁)) = 2 + 2·x₁
POL(p_out_ga(x₁)) = 2·x₁

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ PisEmptyProof

Q DP problem:
P is empty.
R is empty.
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.