Left Termination proof of /tmp/tmpn3agJ3/rotate.pl

Left Termination of the query pattern rotate_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:

↳ Prolog

  ↳ PrologToPiTRSProof

Clauses:

rotate(X, Y) :- ','(append2(A, B, X), append1(B, A, Y)).
append1(.(X, Xs), Ys, .(X, Zs)) :- append1(Xs, Ys, Zs).
append1([], Ys, Ys).
append2(.(X, Xs), Ys, .(X, Zs)) :- append2(Xs, Ys, Zs).
append2([], Ys, Ys).

Queries:

rotate(g,a).

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:

rotate_in(X, Y) → U1(X, Y, append2_in(A, B, X))
append2_in([], Ys, Ys) → append2_out([], Ys, Ys)
append2_in(.(X, Xs), Ys, .(X, Zs)) → U4(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
U4(X, Xs, Ys, Zs, append2_out(Xs, Ys, Zs)) → append2_out(.(X, Xs), Ys, .(X, Zs))
U1(X, Y, append2_out(A, B, X)) → U2(X, Y, append1_in(B, A, Y))
append1_in([], Ys, Ys) → append1_out([], Ys, Ys)
append1_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
U3(X, Xs, Ys, Zs, append1_out(Xs, Ys, Zs)) → append1_out(.(X, Xs), Ys, .(X, Zs))
U2(X, Y, append1_out(B, A, Y)) → rotate_out(X, 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:

rotate_in(X, Y) → U1(X, Y, append2_in(A, B, X))
append2_in([], Ys, Ys) → append2_out([], Ys, Ys)
append2_in(.(X, Xs), Ys, .(X, Zs)) → U4(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
U4(X, Xs, Ys, Zs, append2_out(Xs, Ys, Zs)) → append2_out(.(X, Xs), Ys, .(X, Zs))
U1(X, Y, append2_out(A, B, X)) → U2(X, Y, append1_in(B, A, Y))
append1_in([], Ys, Ys) → append1_out([], Ys, Ys)
append1_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
U3(X, Xs, Ys, Zs, append1_out(Xs, Ys, Zs)) → append1_out(.(X, Xs), Ys, .(X, Zs))
U2(X, Y, append1_out(B, A, Y)) → rotate_out(X, Y)

The argument filtering Pi contains the following mapping:
rotate_in(x1, x2) = rotate_in(x1)
U1(x1, x2, x3) = U1(x3)
append2_in(x1, x2, x3) = append2_in(x3)
[] = []
append2_out(x1, x2, x3) = append2_out(x1, x2)
.(x1, x2) = .(x1, x2)
U4(x1, x2, x3, x4, x5) = U4(x1, x5)
U2(x1, x2, x3) = U2(x3)
append1_in(x1, x2, x3) = append1_in(x1, x2)
append1_out(x1, x2, x3) = append1_out(x3)
U3(x1, x2, x3, x4, x5) = U3(x1, x5)
rotate_out(x1, x2) = rotate_out(x2)

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

ROTATE_IN(X, Y) → U1¹(X, Y, append2_in(A, B, X))
ROTATE_IN(X, Y) → APPEND2_IN(A, B, X)
APPEND2_IN(.(X, Xs), Ys, .(X, Zs)) → U4¹(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
APPEND2_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND2_IN(Xs, Ys, Zs)
U1¹(X, Y, append2_out(A, B, X)) → U2¹(X, Y, append1_in(B, A, Y))
U1¹(X, Y, append2_out(A, B, X)) → APPEND1_IN(B, A, Y)
APPEND1_IN(.(X, Xs), Ys, .(X, Zs)) → U3¹(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
APPEND1_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND1_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

rotate_in(X, Y) → U1(X, Y, append2_in(A, B, X))
append2_in([], Ys, Ys) → append2_out([], Ys, Ys)
append2_in(.(X, Xs), Ys, .(X, Zs)) → U4(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
U4(X, Xs, Ys, Zs, append2_out(Xs, Ys, Zs)) → append2_out(.(X, Xs), Ys, .(X, Zs))
U1(X, Y, append2_out(A, B, X)) → U2(X, Y, append1_in(B, A, Y))
append1_in([], Ys, Ys) → append1_out([], Ys, Ys)
append1_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
U3(X, Xs, Ys, Zs, append1_out(Xs, Ys, Zs)) → append1_out(.(X, Xs), Ys, .(X, Zs))
U2(X, Y, append1_out(B, A, Y)) → rotate_out(X, Y)

The argument filtering Pi contains the following mapping:
rotate_in(x1, x2) = rotate_in(x1)
U1(x1, x2, x3) = U1(x3)
append2_in(x1, x2, x3) = append2_in(x3)
[] = []
append2_out(x1, x2, x3) = append2_out(x1, x2)
.(x1, x2) = .(x1, x2)
U4(x1, x2, x3, x4, x5) = U4(x1, x5)
U2(x1, x2, x3) = U2(x3)
append1_in(x1, x2, x3) = append1_in(x1, x2)
append1_out(x1, x2, x3) = append1_out(x3)
U3(x1, x2, x3, x4, x5) = U3(x1, x5)
rotate_out(x1, x2) = rotate_out(x2)
APPEND1_IN(x1, x2, x3) = APPEND1_IN(x1, x2)
U3¹(x1, x2, x3, x4, x5) = U3¹(x1, x5)
U4¹(x1, x2, x3, x4, x5) = U4¹(x1, x5)
APPEND2_IN(x1, x2, x3) = APPEND2_IN(x3)
U2¹(x1, x2, x3) = U2¹(x3)
U1¹(x1, x2, x3) = U1¹(x3)
ROTATE_IN(x1, x2) = ROTATE_IN(x1)

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:

ROTATE_IN(X, Y) → U1¹(X, Y, append2_in(A, B, X))
ROTATE_IN(X, Y) → APPEND2_IN(A, B, X)
APPEND2_IN(.(X, Xs), Ys, .(X, Zs)) → U4¹(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
APPEND2_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND2_IN(Xs, Ys, Zs)
U1¹(X, Y, append2_out(A, B, X)) → U2¹(X, Y, append1_in(B, A, Y))
U1¹(X, Y, append2_out(A, B, X)) → APPEND1_IN(B, A, Y)
APPEND1_IN(.(X, Xs), Ys, .(X, Zs)) → U3¹(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
APPEND1_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND1_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

rotate_in(X, Y) → U1(X, Y, append2_in(A, B, X))
append2_in([], Ys, Ys) → append2_out([], Ys, Ys)
append2_in(.(X, Xs), Ys, .(X, Zs)) → U4(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
U4(X, Xs, Ys, Zs, append2_out(Xs, Ys, Zs)) → append2_out(.(X, Xs), Ys, .(X, Zs))
U1(X, Y, append2_out(A, B, X)) → U2(X, Y, append1_in(B, A, Y))
append1_in([], Ys, Ys) → append1_out([], Ys, Ys)
append1_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
U3(X, Xs, Ys, Zs, append1_out(Xs, Ys, Zs)) → append1_out(.(X, Xs), Ys, .(X, Zs))
U2(X, Y, append1_out(B, A, Y)) → rotate_out(X, Y)

The argument filtering Pi contains the following mapping:
rotate_in(x1, x2) = rotate_in(x1)
U1(x1, x2, x3) = U1(x3)
append2_in(x1, x2, x3) = append2_in(x3)
[] = []
append2_out(x1, x2, x3) = append2_out(x1, x2)
.(x1, x2) = .(x1, x2)
U4(x1, x2, x3, x4, x5) = U4(x1, x5)
U2(x1, x2, x3) = U2(x3)
append1_in(x1, x2, x3) = append1_in(x1, x2)
append1_out(x1, x2, x3) = append1_out(x3)
U3(x1, x2, x3, x4, x5) = U3(x1, x5)
rotate_out(x1, x2) = rotate_out(x2)
APPEND1_IN(x1, x2, x3) = APPEND1_IN(x1, x2)
U3¹(x1, x2, x3, x4, x5) = U3¹(x1, x5)
U4¹(x1, x2, x3, x4, x5) = U4¹(x1, x5)
APPEND2_IN(x1, x2, x3) = APPEND2_IN(x3)
U2¹(x1, x2, x3) = U2¹(x3)
U1¹(x1, x2, x3) = U1¹(x3)
ROTATE_IN(x1, x2) = ROTATE_IN(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 6 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

APPEND1_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND1_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

rotate_in(X, Y) → U1(X, Y, append2_in(A, B, X))
append2_in([], Ys, Ys) → append2_out([], Ys, Ys)
append2_in(.(X, Xs), Ys, .(X, Zs)) → U4(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
U4(X, Xs, Ys, Zs, append2_out(Xs, Ys, Zs)) → append2_out(.(X, Xs), Ys, .(X, Zs))
U1(X, Y, append2_out(A, B, X)) → U2(X, Y, append1_in(B, A, Y))
append1_in([], Ys, Ys) → append1_out([], Ys, Ys)
append1_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
U3(X, Xs, Ys, Zs, append1_out(Xs, Ys, Zs)) → append1_out(.(X, Xs), Ys, .(X, Zs))
U2(X, Y, append1_out(B, A, Y)) → rotate_out(X, Y)

The argument filtering Pi contains the following mapping:
rotate_in(x1, x2) = rotate_in(x1)
U1(x1, x2, x3) = U1(x3)
append2_in(x1, x2, x3) = append2_in(x3)
[] = []
append2_out(x1, x2, x3) = append2_out(x1, x2)
.(x1, x2) = .(x1, x2)
U4(x1, x2, x3, x4, x5) = U4(x1, x5)
U2(x1, x2, x3) = U2(x3)
append1_in(x1, x2, x3) = append1_in(x1, x2)
append1_out(x1, x2, x3) = append1_out(x3)
U3(x1, x2, x3, x4, x5) = U3(x1, x5)
rotate_out(x1, x2) = rotate_out(x2)
APPEND1_IN(x1, x2, x3) = APPEND1_IN(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:

APPEND1_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND1_IN(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
APPEND1_IN(x1, x2, x3) = APPEND1_IN(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

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

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

APPEND1_IN(.(X, Xs), Ys) → APPEND1_IN(Xs, Ys)

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:

APPEND1_IN(.(X, Xs), Ys) → APPEND1_IN(Xs, Ys)
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:

APPEND2_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND2_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

rotate_in(X, Y) → U1(X, Y, append2_in(A, B, X))
append2_in([], Ys, Ys) → append2_out([], Ys, Ys)
append2_in(.(X, Xs), Ys, .(X, Zs)) → U4(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
U4(X, Xs, Ys, Zs, append2_out(Xs, Ys, Zs)) → append2_out(.(X, Xs), Ys, .(X, Zs))
U1(X, Y, append2_out(A, B, X)) → U2(X, Y, append1_in(B, A, Y))
append1_in([], Ys, Ys) → append1_out([], Ys, Ys)
append1_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
U3(X, Xs, Ys, Zs, append1_out(Xs, Ys, Zs)) → append1_out(.(X, Xs), Ys, .(X, Zs))
U2(X, Y, append1_out(B, A, Y)) → rotate_out(X, Y)

The argument filtering Pi contains the following mapping:
rotate_in(x1, x2) = rotate_in(x1)
U1(x1, x2, x3) = U1(x3)
append2_in(x1, x2, x3) = append2_in(x3)
[] = []
append2_out(x1, x2, x3) = append2_out(x1, x2)
.(x1, x2) = .(x1, x2)
U4(x1, x2, x3, x4, x5) = U4(x1, x5)
U2(x1, x2, x3) = U2(x3)
append1_in(x1, x2, x3) = append1_in(x1, x2)
append1_out(x1, x2, x3) = append1_out(x3)
U3(x1, x2, x3, x4, x5) = U3(x1, x5)
rotate_out(x1, x2) = rotate_out(x2)
APPEND2_IN(x1, x2, x3) = APPEND2_IN(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

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

APPEND2_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND2_IN(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
APPEND2_IN(x1, x2, x3) = APPEND2_IN(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

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

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

APPEND2_IN(.(X, Zs)) → APPEND2_IN(Zs)

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

APPEND2_IN(.(X, Zs)) → APPEND2_IN(Zs)
The graph contains the following edges 1 > 1