Left Termination proof of /tmp/tmp1YXlDL/p.pl

Left Termination of the query pattern p_in_1(g) w.r.t. the given Prolog program could successfully be proven:

↳ Prolog

  ↳ PrologToPiTRSProof

Clauses:

p(cons(X, nil)).
p(cons(s(s(X)), cons(Y, Xs))) :- ','(p(cons(X, cons(Y, Xs))), p(cons(s(s(s(s(Y)))), Xs))).
p(cons(0, Xs)) :- p(Xs).

Queries:

p(g).

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

p_in_g(cons(X, nil)) → p_out_g(cons(X, nil))
p_in_g(cons(s(s(X)), cons(Y, Xs))) → U1_g(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
p_in_g(cons(0, Xs)) → U3_g(Xs, p_in_g(Xs))
U3_g(Xs, p_out_g(Xs)) → p_out_g(cons(0, Xs))
U1_g(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → U2_g(X, Y, Xs, p_in_g(cons(s(s(s(s(Y)))), Xs)))
U2_g(X, Y, Xs, p_out_g(cons(s(s(s(s(Y)))), Xs))) → p_out_g(cons(s(s(X)), cons(Y, Xs)))

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_g(cons(X, nil)) → p_out_g(cons(X, nil))
p_in_g(cons(s(s(X)), cons(Y, Xs))) → U1_g(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
p_in_g(cons(0, Xs)) → U3_g(Xs, p_in_g(Xs))
U3_g(Xs, p_out_g(Xs)) → p_out_g(cons(0, Xs))
U1_g(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → U2_g(X, Y, Xs, p_in_g(cons(s(s(s(s(Y)))), Xs)))
U2_g(X, Y, Xs, p_out_g(cons(s(s(s(s(Y)))), Xs))) → p_out_g(cons(s(s(X)), cons(Y, Xs)))

The argument filtering Pi contains the following mapping:
p_in_g(x1) = p_in_g(x1)
cons(x1, x2) = cons(x1, x2)
nil = nil
p_out_g(x1) = p_out_g
s(x1) = s(x1)
U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4)
0 = 0
U3_g(x1, x2) = U3_g(x2)
U2_g(x1, x2, x3, x4) = U2_g(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_G(cons(s(s(X)), cons(Y, Xs))) → U1_G(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
P_IN_G(cons(s(s(X)), cons(Y, Xs))) → P_IN_G(cons(X, cons(Y, Xs)))
P_IN_G(cons(0, Xs)) → U3_G(Xs, p_in_g(Xs))
P_IN_G(cons(0, Xs)) → P_IN_G(Xs)
U1_G(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → U2_G(X, Y, Xs, p_in_g(cons(s(s(s(s(Y)))), Xs)))
U1_G(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → P_IN_G(cons(s(s(s(s(Y)))), Xs))

The TRS R consists of the following rules:

p_in_g(cons(X, nil)) → p_out_g(cons(X, nil))
p_in_g(cons(s(s(X)), cons(Y, Xs))) → U1_g(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
p_in_g(cons(0, Xs)) → U3_g(Xs, p_in_g(Xs))
U3_g(Xs, p_out_g(Xs)) → p_out_g(cons(0, Xs))
U1_g(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → U2_g(X, Y, Xs, p_in_g(cons(s(s(s(s(Y)))), Xs)))
U2_g(X, Y, Xs, p_out_g(cons(s(s(s(s(Y)))), Xs))) → p_out_g(cons(s(s(X)), cons(Y, Xs)))

The argument filtering Pi contains the following mapping:
p_in_g(x1) = p_in_g(x1)
cons(x1, x2) = cons(x1, x2)
nil = nil
p_out_g(x1) = p_out_g
s(x1) = s(x1)
U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4)
0 = 0
U3_g(x1, x2) = U3_g(x2)
U2_g(x1, x2, x3, x4) = U2_g(x4)
P_IN_G(x1) = P_IN_G(x1)
U1_G(x1, x2, x3, x4) = U1_G(x2, x3, x4)
U3_G(x1, x2) = U3_G(x2)
U2_G(x1, x2, x3, x4) = U2_G(x4)

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_G(cons(s(s(X)), cons(Y, Xs))) → U1_G(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
P_IN_G(cons(s(s(X)), cons(Y, Xs))) → P_IN_G(cons(X, cons(Y, Xs)))
P_IN_G(cons(0, Xs)) → U3_G(Xs, p_in_g(Xs))
P_IN_G(cons(0, Xs)) → P_IN_G(Xs)
U1_G(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → U2_G(X, Y, Xs, p_in_g(cons(s(s(s(s(Y)))), Xs)))
U1_G(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → P_IN_G(cons(s(s(s(s(Y)))), Xs))

The TRS R consists of the following rules:

p_in_g(cons(X, nil)) → p_out_g(cons(X, nil))
p_in_g(cons(s(s(X)), cons(Y, Xs))) → U1_g(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
p_in_g(cons(0, Xs)) → U3_g(Xs, p_in_g(Xs))
U3_g(Xs, p_out_g(Xs)) → p_out_g(cons(0, Xs))
U1_g(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → U2_g(X, Y, Xs, p_in_g(cons(s(s(s(s(Y)))), Xs)))
U2_g(X, Y, Xs, p_out_g(cons(s(s(s(s(Y)))), Xs))) → p_out_g(cons(s(s(X)), cons(Y, Xs)))

The argument filtering Pi contains the following mapping:
p_in_g(x1) = p_in_g(x1)
cons(x1, x2) = cons(x1, x2)
nil = nil
p_out_g(x1) = p_out_g
s(x1) = s(x1)
U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4)
0 = 0
U3_g(x1, x2) = U3_g(x2)
U2_g(x1, x2, x3, x4) = U2_g(x4)
P_IN_G(x1) = P_IN_G(x1)
U1_G(x1, x2, x3, x4) = U1_G(x2, x3, x4)
U3_G(x1, x2) = U3_G(x2)
U2_G(x1, x2, x3, x4) = U2_G(x4)

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

              ↳ PiDPToQDPProof

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

P_IN_G(cons(0, Xs)) → P_IN_G(Xs)
P_IN_G(cons(s(s(X)), cons(Y, Xs))) → P_IN_G(cons(X, cons(Y, Xs)))
P_IN_G(cons(s(s(X)), cons(Y, Xs))) → U1_G(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
U1_G(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → P_IN_G(cons(s(s(s(s(Y)))), Xs))

The TRS R consists of the following rules:

p_in_g(cons(X, nil)) → p_out_g(cons(X, nil))
p_in_g(cons(s(s(X)), cons(Y, Xs))) → U1_g(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
p_in_g(cons(0, Xs)) → U3_g(Xs, p_in_g(Xs))
U3_g(Xs, p_out_g(Xs)) → p_out_g(cons(0, Xs))
U1_g(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → U2_g(X, Y, Xs, p_in_g(cons(s(s(s(s(Y)))), Xs)))
U2_g(X, Y, Xs, p_out_g(cons(s(s(s(s(Y)))), Xs))) → p_out_g(cons(s(s(X)), cons(Y, Xs)))

The argument filtering Pi contains the following mapping:
p_in_g(x1) = p_in_g(x1)
cons(x1, x2) = cons(x1, x2)
nil = nil
p_out_g(x1) = p_out_g
s(x1) = s(x1)
U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4)
0 = 0
U3_g(x1, x2) = U3_g(x2)
U2_g(x1, x2, x3, x4) = U2_g(x4)
P_IN_G(x1) = P_IN_G(x1)
U1_G(x1, x2, x3, x4) = U1_G(x2, x3, x4)

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

              ↳ PiDPToQDPProof

                ↳ QDP

                  ↳ QDPOrderProof

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

P_IN_G(cons(0, Xs)) → P_IN_G(Xs)
P_IN_G(cons(s(s(X)), cons(Y, Xs))) → U1_G(Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
P_IN_G(cons(s(s(X)), cons(Y, Xs))) → P_IN_G(cons(X, cons(Y, Xs)))
U1_G(Y, Xs, p_out_g) → P_IN_G(cons(s(s(s(s(Y)))), Xs))

The TRS R consists of the following rules:

p_in_g(cons(X, nil)) → p_out_g
p_in_g(cons(s(s(X)), cons(Y, Xs))) → U1_g(Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
p_in_g(cons(0, Xs)) → U3_g(p_in_g(Xs))
U3_g(p_out_g) → p_out_g
U1_g(Y, Xs, p_out_g) → U2_g(p_in_g(cons(s(s(s(s(Y)))), Xs)))
U2_g(p_out_g) → p_out_g

The set Q consists of the following terms:

p_in_g(x0)
U3_g(x0)
U1_g(x0, x1, x2)
U2_g(x0)

We have to consider all (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

P_IN_G(cons(0, Xs)) → P_IN_G(Xs)

The remaining pairs can at least be oriented weakly.

P_IN_G(cons(s(s(X)), cons(Y, Xs))) → U1_G(Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
P_IN_G(cons(s(s(X)), cons(Y, Xs))) → P_IN_G(cons(X, cons(Y, Xs)))
U1_G(Y, Xs, p_out_g) → P_IN_G(cons(s(s(s(s(Y)))), Xs))

Used ordering: Polynomial interpretation [25]:

POL(0) = 1
POL(P_IN_G(x₁)) = x₁
POL(U1_G(x₁, x₂, x₃)) = x₁ + x₂
POL(U1_g(x₁, x₂, x₃)) = 0
POL(U2_g(x₁)) = 0
POL(U3_g(x₁)) = 0
POL(cons(x₁, x₂)) = x₁ + x₂
POL(nil) = 0
POL(p_in_g(x₁)) = 0
POL(p_out_g) = 0
POL(s(x₁)) = x₁

The following usable rules [17] were oriented: none

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ PiDPToQDPProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPOrderProof

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

P_IN_G(cons(s(s(X)), cons(Y, Xs))) → U1_G(Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
P_IN_G(cons(s(s(X)), cons(Y, Xs))) → P_IN_G(cons(X, cons(Y, Xs)))
U1_G(Y, Xs, p_out_g) → P_IN_G(cons(s(s(s(s(Y)))), Xs))

The TRS R consists of the following rules:

p_in_g(cons(X, nil)) → p_out_g
p_in_g(cons(s(s(X)), cons(Y, Xs))) → U1_g(Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
p_in_g(cons(0, Xs)) → U3_g(p_in_g(Xs))
U3_g(p_out_g) → p_out_g
U1_g(Y, Xs, p_out_g) → U2_g(p_in_g(cons(s(s(s(s(Y)))), Xs)))
U2_g(p_out_g) → p_out_g

The set Q consists of the following terms:

p_in_g(x0)
U3_g(x0)
U1_g(x0, x1, x2)
U2_g(x0)

We have to consider all (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

P_IN_G(cons(s(s(X)), cons(Y, Xs))) → U1_G(Y, Xs, p_in_g(cons(X, cons(Y, Xs))))

The remaining pairs can at least be oriented weakly.

P_IN_G(cons(s(s(X)), cons(Y, Xs))) → P_IN_G(cons(X, cons(Y, Xs)))
U1_G(Y, Xs, p_out_g) → P_IN_G(cons(s(s(s(s(Y)))), Xs))

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(P_IN_G(x₁)) = x₁
POL(U1_G(x₁, x₂, x₃)) = 1 + x₂
POL(U1_g(x₁, x₂, x₃)) = 0
POL(U2_g(x₁)) = 0
POL(U3_g(x₁)) = 0
POL(cons(x₁, x₂)) = 1 + x₂
POL(nil) = 0
POL(p_in_g(x₁)) = x₁
POL(p_out_g) = 0
POL(s(x₁)) = 0

The following usable rules [17] were oriented: none

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ PiDPToQDPProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ DependencyGraphProof

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

P_IN_G(cons(s(s(X)), cons(Y, Xs))) → P_IN_G(cons(X, cons(Y, Xs)))
U1_G(Y, Xs, p_out_g) → P_IN_G(cons(s(s(s(s(Y)))), Xs))

The TRS R consists of the following rules:

p_in_g(cons(X, nil)) → p_out_g
p_in_g(cons(s(s(X)), cons(Y, Xs))) → U1_g(Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
p_in_g(cons(0, Xs)) → U3_g(p_in_g(Xs))
U3_g(p_out_g) → p_out_g
U1_g(Y, Xs, p_out_g) → U2_g(p_in_g(cons(s(s(s(s(Y)))), Xs)))
U2_g(p_out_g) → p_out_g

The set Q consists of the following terms:

p_in_g(x0)
U3_g(x0)
U1_g(x0, x1, x2)
U2_g(x0)

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

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ PiDPToQDPProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesProof

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

P_IN_G(cons(s(s(X)), cons(Y, Xs))) → P_IN_G(cons(X, cons(Y, Xs)))

The TRS R consists of the following rules:

p_in_g(cons(X, nil)) → p_out_g
p_in_g(cons(s(s(X)), cons(Y, Xs))) → U1_g(Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
p_in_g(cons(0, Xs)) → U3_g(p_in_g(Xs))
U3_g(p_out_g) → p_out_g
U1_g(Y, Xs, p_out_g) → U2_g(p_in_g(cons(s(s(s(s(Y)))), Xs)))
U2_g(p_out_g) → p_out_g

The set Q consists of the following terms:

p_in_g(x0)
U3_g(x0)
U1_g(x0, x1, x2)
U2_g(x0)

We have to consider all (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ PiDPToQDPProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesProof

                                ↳ QDP

                                  ↳ QReductionProof

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

P_IN_G(cons(s(s(X)), cons(Y, Xs))) → P_IN_G(cons(X, cons(Y, Xs)))

R is empty.
The set Q consists of the following terms:

p_in_g(x0)
U3_g(x0)
U1_g(x0, x1, x2)
U2_g(x0)

We have to consider all (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

p_in_g(x0)
U3_g(x0)
U1_g(x0, x1, x2)
U2_g(x0)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ PiDPToQDPProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesProof

                                ↳ QDP

                                  ↳ QReductionProof

                                    ↳ QDP

                                      ↳ UsableRulesReductionPairsProof

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

P_IN_G(cons(s(s(X)), cons(Y, Xs))) → P_IN_G(cons(X, cons(Y, Xs)))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

P_IN_G(cons(s(s(X)), cons(Y, Xs))) → P_IN_G(cons(X, cons(Y, Xs)))

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(P_IN_G(x₁)) = 2·x₁
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(s(x₁)) = 2·x₁

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ PiDPToQDPProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesProof

                                ↳ QDP

                                  ↳ QReductionProof

                                    ↳ QDP

                                      ↳ UsableRulesReductionPairsProof

                                        ↳ QDP

                                          ↳ PisEmptyProof

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