Left Termination proof of /tmp/tmpLoaIWN/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].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

p_in(cons(0, Xs)) → U3(Xs, p_in(Xs))
p_in(cons(s(s(X)), cons(Y, Xs))) → U1(X, Y, Xs, p_in(cons(X, cons(Y, Xs))))
p_in(cons(X, nil)) → p_out(cons(X, nil))
U1(X, Y, Xs, p_out(cons(X, cons(Y, Xs)))) → U2(X, Y, Xs, p_in(cons(s(s(s(s(Y)))), Xs)))
U2(X, Y, Xs, p_out(cons(s(s(s(s(Y)))), Xs))) → p_out(cons(s(s(X)), cons(Y, Xs)))
U3(Xs, p_out(Xs)) → p_out(cons(0, 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(cons(0, Xs)) → U3(Xs, p_in(Xs))
p_in(cons(s(s(X)), cons(Y, Xs))) → U1(X, Y, Xs, p_in(cons(X, cons(Y, Xs))))
p_in(cons(X, nil)) → p_out(cons(X, nil))
U1(X, Y, Xs, p_out(cons(X, cons(Y, Xs)))) → U2(X, Y, Xs, p_in(cons(s(s(s(s(Y)))), Xs)))
U2(X, Y, Xs, p_out(cons(s(s(s(s(Y)))), Xs))) → p_out(cons(s(s(X)), cons(Y, Xs)))
U3(Xs, p_out(Xs)) → p_out(cons(0, Xs))

The argument filtering Pi contains the following mapping:
p_in(x1) = p_in(x1)
cons(x1, x2) = cons(x1, x2)
0 = 0
U3(x1, x2) = U3(x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4) = U1(x2, x3, x4)
nil = nil
p_out(x1) = p_out
U2(x1, x2, x3, x4) = U2(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(cons(0, Xs)) → U3¹(Xs, p_in(Xs))
P_IN(cons(0, Xs)) → P_IN(Xs)
P_IN(cons(s(s(X)), cons(Y, Xs))) → U1¹(X, Y, Xs, p_in(cons(X, cons(Y, Xs))))
P_IN(cons(s(s(X)), cons(Y, Xs))) → P_IN(cons(X, cons(Y, Xs)))
U1¹(X, Y, Xs, p_out(cons(X, cons(Y, Xs)))) → U2¹(X, Y, Xs, p_in(cons(s(s(s(s(Y)))), Xs)))
U1¹(X, Y, Xs, p_out(cons(X, cons(Y, Xs)))) → P_IN(cons(s(s(s(s(Y)))), Xs))

The TRS R consists of the following rules:

p_in(cons(0, Xs)) → U3(Xs, p_in(Xs))
p_in(cons(s(s(X)), cons(Y, Xs))) → U1(X, Y, Xs, p_in(cons(X, cons(Y, Xs))))
p_in(cons(X, nil)) → p_out(cons(X, nil))
U1(X, Y, Xs, p_out(cons(X, cons(Y, Xs)))) → U2(X, Y, Xs, p_in(cons(s(s(s(s(Y)))), Xs)))
U2(X, Y, Xs, p_out(cons(s(s(s(s(Y)))), Xs))) → p_out(cons(s(s(X)), cons(Y, Xs)))
U3(Xs, p_out(Xs)) → p_out(cons(0, Xs))

The argument filtering Pi contains the following mapping:
p_in(x1) = p_in(x1)
cons(x1, x2) = cons(x1, x2)
0 = 0
U3(x1, x2) = U3(x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4) = U1(x2, x3, x4)
nil = nil
p_out(x1) = p_out
U2(x1, x2, x3, x4) = U2(x4)
P_IN(x1) = P_IN(x1)
U2¹(x1, x2, x3, x4) = U2¹(x4)
U3¹(x1, x2) = U3¹(x2)
U1¹(x1, x2, x3, x4) = U1¹(x2, x3, 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(cons(0, Xs)) → U3¹(Xs, p_in(Xs))
P_IN(cons(0, Xs)) → P_IN(Xs)
P_IN(cons(s(s(X)), cons(Y, Xs))) → U1¹(X, Y, Xs, p_in(cons(X, cons(Y, Xs))))
P_IN(cons(s(s(X)), cons(Y, Xs))) → P_IN(cons(X, cons(Y, Xs)))
U1¹(X, Y, Xs, p_out(cons(X, cons(Y, Xs)))) → U2¹(X, Y, Xs, p_in(cons(s(s(s(s(Y)))), Xs)))
U1¹(X, Y, Xs, p_out(cons(X, cons(Y, Xs)))) → P_IN(cons(s(s(s(s(Y)))), Xs))

The TRS R consists of the following rules:

p_in(cons(0, Xs)) → U3(Xs, p_in(Xs))
p_in(cons(s(s(X)), cons(Y, Xs))) → U1(X, Y, Xs, p_in(cons(X, cons(Y, Xs))))
p_in(cons(X, nil)) → p_out(cons(X, nil))
U1(X, Y, Xs, p_out(cons(X, cons(Y, Xs)))) → U2(X, Y, Xs, p_in(cons(s(s(s(s(Y)))), Xs)))
U2(X, Y, Xs, p_out(cons(s(s(s(s(Y)))), Xs))) → p_out(cons(s(s(X)), cons(Y, Xs)))
U3(Xs, p_out(Xs)) → p_out(cons(0, Xs))

The argument filtering Pi contains the following mapping:
p_in(x1) = p_in(x1)
cons(x1, x2) = cons(x1, x2)
0 = 0
U3(x1, x2) = U3(x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4) = U1(x2, x3, x4)
nil = nil
p_out(x1) = p_out
U2(x1, x2, x3, x4) = U2(x4)
P_IN(x1) = P_IN(x1)
U2¹(x1, x2, x3, x4) = U2¹(x4)
U3¹(x1, x2) = U3¹(x2)
U1¹(x1, x2, x3, x4) = U1¹(x2, x3, 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(cons(0, Xs)) → P_IN(Xs)
U1¹(X, Y, Xs, p_out(cons(X, cons(Y, Xs)))) → P_IN(cons(s(s(s(s(Y)))), Xs))
P_IN(cons(s(s(X)), cons(Y, Xs))) → U1¹(X, Y, Xs, p_in(cons(X, cons(Y, Xs))))
P_IN(cons(s(s(X)), cons(Y, Xs))) → P_IN(cons(X, cons(Y, Xs)))

The TRS R consists of the following rules:

p_in(cons(0, Xs)) → U3(Xs, p_in(Xs))
p_in(cons(s(s(X)), cons(Y, Xs))) → U1(X, Y, Xs, p_in(cons(X, cons(Y, Xs))))
p_in(cons(X, nil)) → p_out(cons(X, nil))
U1(X, Y, Xs, p_out(cons(X, cons(Y, Xs)))) → U2(X, Y, Xs, p_in(cons(s(s(s(s(Y)))), Xs)))
U2(X, Y, Xs, p_out(cons(s(s(s(s(Y)))), Xs))) → p_out(cons(s(s(X)), cons(Y, Xs)))
U3(Xs, p_out(Xs)) → p_out(cons(0, Xs))

The argument filtering Pi contains the following mapping:
p_in(x1) = p_in(x1)
cons(x1, x2) = cons(x1, x2)
0 = 0
U3(x1, x2) = U3(x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4) = U1(x2, x3, x4)
nil = nil
p_out(x1) = p_out
U2(x1, x2, x3, x4) = U2(x4)
P_IN(x1) = P_IN(x1)
U1¹(x1, x2, x3, x4) = U1¹(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(cons(0, Xs)) → P_IN(Xs)
U1¹(Y, Xs, p_out) → P_IN(cons(s(s(s(s(Y)))), Xs))
P_IN(cons(s(s(X)), cons(Y, Xs))) → U1¹(Y, Xs, p_in(cons(X, cons(Y, Xs))))
P_IN(cons(s(s(X)), cons(Y, Xs))) → P_IN(cons(X, cons(Y, Xs)))

The TRS R consists of the following rules:

p_in(cons(0, Xs)) → U3(p_in(Xs))
p_in(cons(s(s(X)), cons(Y, Xs))) → U1(Y, Xs, p_in(cons(X, cons(Y, Xs))))
p_in(cons(X, nil)) → p_out
U1(Y, Xs, p_out) → U2(p_in(cons(s(s(s(s(Y)))), Xs)))
U2(p_out) → p_out
U3(p_out) → p_out

The set Q consists of the following terms:

p_in(x0)
U1(x0, x1, x2)
U2(x0)
U3(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(cons(0, Xs)) → P_IN(Xs)
U1¹(Y, Xs, p_out) → P_IN(cons(s(s(s(s(Y)))), Xs))

The remaining pairs can at least be oriented weakly.

P_IN(cons(s(s(X)), cons(Y, Xs))) → U1¹(Y, Xs, p_in(cons(X, cons(Y, Xs))))
P_IN(cons(s(s(X)), cons(Y, Xs))) → P_IN(cons(X, cons(Y, Xs)))

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(P_IN(x₁)) = x₁
POL(U1(x₁, x₂, x₃)) = x₃
POL(U1¹(x₁, x₂, x₃)) = 1 + x₂ + x₃
POL(U2(x₁)) = 1
POL(U3(x₁)) = 1
POL(cons(x₁, x₂)) = 1 + x₂
POL(nil) = 0
POL(p_in(x₁)) = 1
POL(p_out) = 1
POL(s(x₁)) = 0

The following usable rules [17] were oriented:

U3(p_out) → p_out
p_in(cons(s(s(X)), cons(Y, Xs))) → U1(Y, Xs, p_in(cons(X, cons(Y, Xs))))
U1(Y, Xs, p_out) → U2(p_in(cons(s(s(s(s(Y)))), Xs)))
U2(p_out) → p_out
p_in(cons(0, Xs)) → U3(p_in(Xs))

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ PiDPToQDPProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

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

P_IN(cons(s(s(X)), cons(Y, Xs))) → U1¹(Y, Xs, p_in(cons(X, cons(Y, Xs))))
P_IN(cons(s(s(X)), cons(Y, Xs))) → P_IN(cons(X, cons(Y, Xs)))

The TRS R consists of the following rules:

p_in(cons(0, Xs)) → U3(p_in(Xs))
p_in(cons(s(s(X)), cons(Y, Xs))) → U1(Y, Xs, p_in(cons(X, cons(Y, Xs))))
p_in(cons(X, nil)) → p_out
U1(Y, Xs, p_out) → U2(p_in(cons(s(s(s(s(Y)))), Xs)))
U2(p_out) → p_out
U3(p_out) → p_out

The set Q consists of the following terms:

p_in(x0)
U1(x0, x1, x2)
U2(x0)
U3(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

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

p_in(cons(0, Xs)) → U3(p_in(Xs))
p_in(cons(s(s(X)), cons(Y, Xs))) → U1(Y, Xs, p_in(cons(X, cons(Y, Xs))))
p_in(cons(X, nil)) → p_out
U1(Y, Xs, p_out) → U2(p_in(cons(s(s(s(s(Y)))), Xs)))
U2(p_out) → p_out
U3(p_out) → p_out

The set Q consists of the following terms:

p_in(x0)
U1(x0, x1, x2)
U2(x0)
U3(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

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ UsableRulesProof

                            ↳ QDP

                              ↳ QReductionProof

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

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

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

p_in(x0)
U1(x0, x1, x2)
U2(x0)
U3(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(x0)
U1(x0, x1, x2)
U2(x0)
U3(x0)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ PiDPToQDPProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ UsableRulesProof

                            ↳ QDP

                              ↳ QReductionProof

                                ↳ QDP

                                  ↳ UsableRulesReductionPairsProof

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

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

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(P_IN(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

                      ↳ 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.