Left Termination proof of /tmp/tmprM3xGp/flatlength-bbf.pl

Left Termination of the query pattern fl_in_3(g, g, a) w.r.t. the given Prolog program could successfully be proven:

↳ Prolog

  ↳ PrologToPiTRSProof

Clauses:

fl([], [], 0).
fl(.(E, X), R, s(Z)) :- ','(append(E, Y, R), fl(X, Y, Z)).
append([], X, X).
append(.(X, Xs), Ys, .(X, Zs)) :- append(Xs, Ys, Zs).

Queries:

fl(g,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:

fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))

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:

fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3) = fl_in(x1, x2)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x2, x5)
append_in(x1, x2, x3) = append_in(x1, x3)
U3(x1, x2, x3, x4, x5) = U3(x5)
[] = []
append_out(x1, x2, x3) = append_out(x2)
U2(x1, x2, x3, x4, x5) = U2(x5)
0 = 0
fl_out(x1, x2, x3) = fl_out(x3)

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

FL_IN(.(E, X), R, s(Z)) → U1¹(E, X, R, Z, append_in(E, Y, R))
FL_IN(.(E, X), R, s(Z)) → APPEND_IN(E, Y, R)
APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → U3¹(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN(Xs, Ys, Zs)
U1¹(E, X, R, Z, append_out(E, Y, R)) → U2¹(E, X, R, Z, fl_in(X, Y, Z))
U1¹(E, X, R, Z, append_out(E, Y, R)) → FL_IN(X, Y, Z)

The TRS R consists of the following rules:

fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3) = fl_in(x1, x2)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x2, x5)
append_in(x1, x2, x3) = append_in(x1, x3)
U3(x1, x2, x3, x4, x5) = U3(x5)
[] = []
append_out(x1, x2, x3) = append_out(x2)
U2(x1, x2, x3, x4, x5) = U2(x5)
0 = 0
fl_out(x1, x2, x3) = fl_out(x3)
U3¹(x1, x2, x3, x4, x5) = U3¹(x5)
APPEND_IN(x1, x2, x3) = APPEND_IN(x1, x3)
FL_IN(x1, x2, x3) = FL_IN(x1, x2)
U2¹(x1, x2, x3, x4, x5) = U2¹(x5)
U1¹(x1, x2, x3, x4, x5) = U1¹(x2, x5)

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:

FL_IN(.(E, X), R, s(Z)) → U1¹(E, X, R, Z, append_in(E, Y, R))
FL_IN(.(E, X), R, s(Z)) → APPEND_IN(E, Y, R)
APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → U3¹(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN(Xs, Ys, Zs)
U1¹(E, X, R, Z, append_out(E, Y, R)) → U2¹(E, X, R, Z, fl_in(X, Y, Z))
U1¹(E, X, R, Z, append_out(E, Y, R)) → FL_IN(X, Y, Z)

The TRS R consists of the following rules:

fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3) = fl_in(x1, x2)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x2, x5)
append_in(x1, x2, x3) = append_in(x1, x3)
U3(x1, x2, x3, x4, x5) = U3(x5)
[] = []
append_out(x1, x2, x3) = append_out(x2)
U2(x1, x2, x3, x4, x5) = U2(x5)
0 = 0
fl_out(x1, x2, x3) = fl_out(x3)
U3¹(x1, x2, x3, x4, x5) = U3¹(x5)
APPEND_IN(x1, x2, x3) = APPEND_IN(x1, x3)
FL_IN(x1, x2, x3) = FL_IN(x1, x2)
U2¹(x1, x2, x3, x4, x5) = U2¹(x5)
U1¹(x1, x2, x3, x4, x5) = U1¹(x2, x5)

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:

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

The TRS R consists of the following rules:

fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3) = fl_in(x1, x2)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x2, x5)
append_in(x1, x2, x3) = append_in(x1, x3)
U3(x1, x2, x3, x4, x5) = U3(x5)
[] = []
append_out(x1, x2, x3) = append_out(x2)
U2(x1, x2, x3, x4, x5) = U2(x5)
0 = 0
fl_out(x1, x2, x3) = fl_out(x3)
APPEND_IN(x1, x2, x3) = APPEND_IN(x1, 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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

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

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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

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

APPEND_IN(.(X, Xs), .(X, Zs)) → APPEND_IN(Xs, Zs)

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:

APPEND_IN(.(X, Xs), .(X, Zs)) → APPEND_IN(Xs, Zs)
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:

FL_IN(.(E, X), R, s(Z)) → U1¹(E, X, R, Z, append_in(E, Y, R))
U1¹(E, X, R, Z, append_out(E, Y, R)) → FL_IN(X, Y, Z)

The TRS R consists of the following rules:

fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3) = fl_in(x1, x2)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x2, x5)
append_in(x1, x2, x3) = append_in(x1, x3)
U3(x1, x2, x3, x4, x5) = U3(x5)
[] = []
append_out(x1, x2, x3) = append_out(x2)
U2(x1, x2, x3, x4, x5) = U2(x5)
0 = 0
fl_out(x1, x2, x3) = fl_out(x3)
FL_IN(x1, x2, x3) = FL_IN(x1, x2)
U1¹(x1, x2, x3, x4, x5) = U1¹(x2, x5)

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:

FL_IN(.(E, X), R, s(Z)) → U1¹(E, X, R, Z, append_in(E, Y, R))
U1¹(E, X, R, Z, append_out(E, Y, R)) → FL_IN(X, Y, Z)

The TRS R consists of the following rules:

append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))

The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
append_in(x1, x2, x3) = append_in(x1, x3)
U3(x1, x2, x3, x4, x5) = U3(x5)
[] = []
append_out(x1, x2, x3) = append_out(x2)
FL_IN(x1, x2, x3) = FL_IN(x1, x2)
U1¹(x1, x2, x3, x4, x5) = U1¹(x2, x5)

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:

FL_IN(.(E, X), R) → U1¹(X, append_in(E, R))
U1¹(X, append_out(Y)) → FL_IN(X, Y)

The TRS R consists of the following rules:

append_in(.(X, Xs), .(X, Zs)) → U3(append_in(Xs, Zs))
append_in([], X) → append_out(X)
U3(append_out(Ys)) → append_out(Ys)

The set Q consists of the following terms:

append_in(x0, x1)
U3(x0)

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:

U1¹(X, append_out(Y)) → FL_IN(X, Y)
The graph contains the following edges 1 >= 1, 2 > 2
FL_IN(.(E, X), R) → U1¹(X, append_in(E, R))
The graph contains the following edges 1 > 1