Left Termination proof of /tmp/tmpsn5q1-/flatlength-bff.pl

Left Termination of the query pattern fl_in_3(g, a, 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,a,a).

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

fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(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_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_gaa(x1, x2, x3) = fl_in_gaa(x1)
[] = []
fl_out_gaa(x1, x2, x3) = fl_out_gaa(x3)
.(x1, x2) = .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5) = U1_gaa(x2, x5)
append_in_gaa(x1, x2, x3) = append_in_gaa(x1)
append_out_gaa(x1, x2, x3) = append_out_gaa
U3_gaa(x1, x2, x3, x4, x5) = U3_gaa(x5)
U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5)
s(x1) = s(x1)

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_GAA(.(E, X), R, s(Z)) → U1_GAA(E, X, R, Z, append_in_gaa(E, Y, R))
FL_IN_GAA(.(E, X), R, s(Z)) → APPEND_IN_GAA(E, Y, R)
APPEND_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → U3_GAA(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
APPEND_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_GAA(Xs, Ys, Zs)
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_GAA(E, X, R, Z, fl_in_gaa(X, Y, Z))
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → FL_IN_GAA(X, Y, Z)

The TRS R consists of the following rules:

fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_gaa(x1, x2, x3) = fl_in_gaa(x1)
[] = []
fl_out_gaa(x1, x2, x3) = fl_out_gaa(x3)
.(x1, x2) = .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5) = U1_gaa(x2, x5)
append_in_gaa(x1, x2, x3) = append_in_gaa(x1)
append_out_gaa(x1, x2, x3) = append_out_gaa
U3_gaa(x1, x2, x3, x4, x5) = U3_gaa(x5)
U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5)
s(x1) = s(x1)
APPEND_IN_GAA(x1, x2, x3) = APPEND_IN_GAA(x1)
U2_GAA(x1, x2, x3, x4, x5) = U2_GAA(x5)
FL_IN_GAA(x1, x2, x3) = FL_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4, x5) = U3_GAA(x5)
U1_GAA(x1, x2, x3, x4, x5) = U1_GAA(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_GAA(.(E, X), R, s(Z)) → U1_GAA(E, X, R, Z, append_in_gaa(E, Y, R))
FL_IN_GAA(.(E, X), R, s(Z)) → APPEND_IN_GAA(E, Y, R)
APPEND_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → U3_GAA(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
APPEND_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_GAA(Xs, Ys, Zs)
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_GAA(E, X, R, Z, fl_in_gaa(X, Y, Z))
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → FL_IN_GAA(X, Y, Z)

The TRS R consists of the following rules:

fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_gaa(x1, x2, x3) = fl_in_gaa(x1)
[] = []
fl_out_gaa(x1, x2, x3) = fl_out_gaa(x3)
.(x1, x2) = .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5) = U1_gaa(x2, x5)
append_in_gaa(x1, x2, x3) = append_in_gaa(x1)
append_out_gaa(x1, x2, x3) = append_out_gaa
U3_gaa(x1, x2, x3, x4, x5) = U3_gaa(x5)
U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5)
s(x1) = s(x1)
APPEND_IN_GAA(x1, x2, x3) = APPEND_IN_GAA(x1)
U2_GAA(x1, x2, x3, x4, x5) = U2_GAA(x5)
FL_IN_GAA(x1, x2, x3) = FL_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4, x5) = U3_GAA(x5)
U1_GAA(x1, x2, x3, x4, x5) = U1_GAA(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_GAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_GAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_gaa(x1, x2, x3) = fl_in_gaa(x1)
[] = []
fl_out_gaa(x1, x2, x3) = fl_out_gaa(x3)
.(x1, x2) = .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5) = U1_gaa(x2, x5)
append_in_gaa(x1, x2, x3) = append_in_gaa(x1)
append_out_gaa(x1, x2, x3) = append_out_gaa
U3_gaa(x1, x2, x3, x4, x5) = U3_gaa(x5)
U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5)
s(x1) = s(x1)
APPEND_IN_GAA(x1, x2, x3) = APPEND_IN_GAA(x1)

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_GAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_GAA(Xs, Ys, Zs)

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

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_GAA(.(X, Xs)) → APPEND_IN_GAA(Xs)

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_GAA(.(X, Xs)) → APPEND_IN_GAA(Xs)
The graph contains the following edges 1 > 1

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

FL_IN_GAA(.(E, X), R, s(Z)) → U1_GAA(E, X, R, Z, append_in_gaa(E, Y, R))
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → FL_IN_GAA(X, Y, Z)

The TRS R consists of the following rules:

fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_gaa(x1, x2, x3) = fl_in_gaa(x1)
[] = []
fl_out_gaa(x1, x2, x3) = fl_out_gaa(x3)
.(x1, x2) = .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5) = U1_gaa(x2, x5)
append_in_gaa(x1, x2, x3) = append_in_gaa(x1)
append_out_gaa(x1, x2, x3) = append_out_gaa
U3_gaa(x1, x2, x3, x4, x5) = U3_gaa(x5)
U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5)
s(x1) = s(x1)
FL_IN_GAA(x1, x2, x3) = FL_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5) = U1_GAA(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_GAA(.(E, X), R, s(Z)) → U1_GAA(E, X, R, Z, append_in_gaa(E, Y, R))
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → FL_IN_GAA(X, Y, Z)

The TRS R consists of the following rules:

append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))

The argument filtering Pi contains the following mapping:
[] = []
.(x1, x2) = .(x1, x2)
append_in_gaa(x1, x2, x3) = append_in_gaa(x1)
append_out_gaa(x1, x2, x3) = append_out_gaa
U3_gaa(x1, x2, x3, x4, x5) = U3_gaa(x5)
s(x1) = s(x1)
FL_IN_GAA(x1, x2, x3) = FL_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5) = U1_GAA(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_GAA(.(E, X)) → U1_GAA(X, append_in_gaa(E))
U1_GAA(X, append_out_gaa) → FL_IN_GAA(X)

The TRS R consists of the following rules:

append_in_gaa([]) → append_out_gaa
append_in_gaa(.(X, Xs)) → U3_gaa(append_in_gaa(Xs))
U3_gaa(append_out_gaa) → append_out_gaa

The set Q consists of the following terms:

append_in_gaa(x0)
U3_gaa(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:

U1_GAA(X, append_out_gaa) → FL_IN_GAA(X)
The graph contains the following edges 1 >= 1
FL_IN_GAA(.(E, X)) → U1_GAA(X, append_in_gaa(E))
The graph contains the following edges 1 > 1