Left Termination proof of /tmp/tmpQmG9oe/ordered.pl

Left Termination of the query pattern ordered_in_1(g) w.r.t. the given Prolog program could successfully be proven:

↳ Prolog

  ↳ PrologToPiTRSProof

Clauses:

ordered([]).
ordered(.(X, [])).
ordered(.(X, .(Y, Xs))) :- ','(le(X, Y), ordered(.(Y, Xs))).
le(s(X), s(Y)) :- le(X, Y).
le(0, s(0)).
le(0, 0).

Queries:

ordered(g).

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

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, le_in_gg(X, Y))
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, le_out_gg(X, Y)) → U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) → ordered_out_g(.(X, .(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:

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, le_in_gg(X, Y))
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, le_out_gg(X, Y)) → U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) → ordered_out_g(.(X, .(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_in_g(x1) = ordered_in_g(x1)
[] = []
ordered_out_g(x1) = ordered_out_g
.(x1, x2) = .(x1, x2)
U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4)
le_in_gg(x1, x2) = le_in_gg(x1, x2)
s(x1) = s(x1)
U3_gg(x1, x2, x3) = U3_gg(x3)
0 = 0
le_out_gg(x1, x2) = le_out_gg
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:

ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(X, Y, Xs, le_in_gg(X, Y))
ORDERED_IN_G(.(X, .(Y, Xs))) → LE_IN_GG(X, Y)
LE_IN_GG(s(X), s(Y)) → U3_GG(X, Y, le_in_gg(X, Y))
LE_IN_GG(s(X), s(Y)) → LE_IN_GG(X, Y)
U1_G(X, Y, Xs, le_out_gg(X, Y)) → U2_G(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U1_G(X, Y, Xs, le_out_gg(X, Y)) → ORDERED_IN_G(.(Y, Xs))

The TRS R consists of the following rules:

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, le_in_gg(X, Y))
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, le_out_gg(X, Y)) → U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) → ordered_out_g(.(X, .(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_in_g(x1) = ordered_in_g(x1)
[] = []
ordered_out_g(x1) = ordered_out_g
.(x1, x2) = .(x1, x2)
U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4)
le_in_gg(x1, x2) = le_in_gg(x1, x2)
s(x1) = s(x1)
U3_gg(x1, x2, x3) = U3_gg(x3)
0 = 0
le_out_gg(x1, x2) = le_out_gg
U2_g(x1, x2, x3, x4) = U2_g(x4)
ORDERED_IN_G(x1) = ORDERED_IN_G(x1)
LE_IN_GG(x1, x2) = LE_IN_GG(x1, x2)
U1_G(x1, x2, x3, x4) = U1_G(x2, x3, x4)
U2_G(x1, x2, x3, x4) = U2_G(x4)
U3_GG(x1, x2, x3) = U3_GG(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:

ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(X, Y, Xs, le_in_gg(X, Y))
ORDERED_IN_G(.(X, .(Y, Xs))) → LE_IN_GG(X, Y)
LE_IN_GG(s(X), s(Y)) → U3_GG(X, Y, le_in_gg(X, Y))
LE_IN_GG(s(X), s(Y)) → LE_IN_GG(X, Y)
U1_G(X, Y, Xs, le_out_gg(X, Y)) → U2_G(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U1_G(X, Y, Xs, le_out_gg(X, Y)) → ORDERED_IN_G(.(Y, Xs))

The TRS R consists of the following rules:

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, le_in_gg(X, Y))
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, le_out_gg(X, Y)) → U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) → ordered_out_g(.(X, .(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_in_g(x1) = ordered_in_g(x1)
[] = []
ordered_out_g(x1) = ordered_out_g
.(x1, x2) = .(x1, x2)
U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4)
le_in_gg(x1, x2) = le_in_gg(x1, x2)
s(x1) = s(x1)
U3_gg(x1, x2, x3) = U3_gg(x3)
0 = 0
le_out_gg(x1, x2) = le_out_gg
U2_g(x1, x2, x3, x4) = U2_g(x4)
ORDERED_IN_G(x1) = ORDERED_IN_G(x1)
LE_IN_GG(x1, x2) = LE_IN_GG(x1, x2)
U1_G(x1, x2, x3, x4) = U1_G(x2, x3, x4)
U2_G(x1, x2, x3, x4) = U2_G(x4)
U3_GG(x1, x2, x3) = U3_GG(x3)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 3 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

LE_IN_GG(s(X), s(Y)) → LE_IN_GG(X, Y)

The TRS R consists of the following rules:

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, le_in_gg(X, Y))
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, le_out_gg(X, Y)) → U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) → ordered_out_g(.(X, .(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_in_g(x1) = ordered_in_g(x1)
[] = []
ordered_out_g(x1) = ordered_out_g
.(x1, x2) = .(x1, x2)
U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4)
le_in_gg(x1, x2) = le_in_gg(x1, x2)
s(x1) = s(x1)
U3_gg(x1, x2, x3) = U3_gg(x3)
0 = 0
le_out_gg(x1, x2) = le_out_gg
U2_g(x1, x2, x3, x4) = U2_g(x4)
LE_IN_GG(x1, x2) = LE_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

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

LE_IN_GG(s(X), s(Y)) → LE_IN_GG(X, Y)

R is empty.
Pi is empty.
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

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

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

LE_IN_GG(s(X), s(Y)) → LE_IN_GG(X, Y)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

LE_IN_GG(s(X), s(Y)) → LE_IN_GG(X, Y)
The graph contains the following edges 1 > 1, 2 > 2

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(X, Y, Xs, le_in_gg(X, Y))
U1_G(X, Y, Xs, le_out_gg(X, Y)) → ORDERED_IN_G(.(Y, Xs))

The TRS R consists of the following rules:

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, le_in_gg(X, Y))
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, le_out_gg(X, Y)) → U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) → ordered_out_g(.(X, .(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_in_g(x1) = ordered_in_g(x1)
[] = []
ordered_out_g(x1) = ordered_out_g
.(x1, x2) = .(x1, x2)
U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4)
le_in_gg(x1, x2) = le_in_gg(x1, x2)
s(x1) = s(x1)
U3_gg(x1, x2, x3) = U3_gg(x3)
0 = 0
le_out_gg(x1, x2) = le_out_gg
U2_g(x1, x2, x3, x4) = U2_g(x4)
ORDERED_IN_G(x1) = ORDERED_IN_G(x1)
U1_G(x1, x2, x3, x4) = U1_G(x2, x3, x4)

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

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(X, Y, Xs, le_in_gg(X, Y))
U1_G(X, Y, Xs, le_out_gg(X, Y)) → ORDERED_IN_G(.(Y, Xs))

The TRS R consists of the following rules:

le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))

The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
le_in_gg(x1, x2) = le_in_gg(x1, x2)
s(x1) = s(x1)
U3_gg(x1, x2, x3) = U3_gg(x3)
0 = 0
le_out_gg(x1, x2) = le_out_gg
ORDERED_IN_G(x1) = ORDERED_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

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

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

ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(Y, Xs, le_in_gg(X, Y))
U1_G(Y, Xs, le_out_gg) → ORDERED_IN_G(.(Y, Xs))

The TRS R consists of the following rules:

le_in_gg(s(X), s(Y)) → U3_gg(le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg
le_in_gg(0, 0) → le_out_gg
U3_gg(le_out_gg) → le_out_gg

The set Q consists of the following terms:

le_in_gg(x0, x1)
U3_gg(x0)

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:

U1_G(Y, Xs, le_out_gg) → ORDERED_IN_G(.(Y, Xs))

The following rules are removed from R:

le_in_gg(s(X), s(Y)) → U3_gg(le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg
le_in_gg(0, 0) → le_out_gg
U3_gg(le_out_gg) → le_out_gg

Used ordering: POLO with Polynomial interpretation [25]:

POL(.(x₁, x₂)) = 2·x₁ + 2·x₂
POL(0) = 1
POL(ORDERED_IN_G(x₁)) = x₁
POL(U1_G(x₁, x₂, x₃)) = 2·x₁ + 2·x₂ + 2·x₃
POL(U3_gg(x₁)) = 1 + 2·x₁
POL(le_in_gg(x₁, x₂)) = x₁ + x₂
POL(le_out_gg) = 2
POL(s(x₁)) = 2 + 2·x₁

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ DependencyGraphProof

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

ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(Y, Xs, le_in_gg(X, Y))

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

le_in_gg(x0, x1)
U3_gg(x0)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.