Left Termination proof of /tmp/tmpAzP2Bh/flatlength-bff.pl

Left Termination of the query pattern fl(b,f,f) w.r.t. the given Prolog program could successfully be proven:

↳ PROLOG

  ↳ PrologToPiTRSProof

fl₃({}₀, {}₀, 0₀).
fl₃(.₂(E, X), R, s₁(Z)) :- append₃(E, Y, R), fl₃(X, Y, Z).
append₃({}₀, X, X).
append₃(.₂(X, Xs), Ys, .₂(X, Zs)) :- append₃(Xs, Ys, Zs).

With regard to the inferred argument filtering the predicates were used in the following modes:
fl₃: (b,f,f)
append₃: (b,f,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

fl_3_in_gaa₃([]_0, []_0, 0_0) -> fl_3_out_gaa₃([]_0, []_0, 0_0)
fl_3_in_gaa₃(._2₂(E, X), R, s_1₁(Z)) -> if_fl_3_in_1_gaa₅(E, X, R, Z, append_3_in_gaa₃(E, Y, R))
append_3_in_gaa₃([]_0, X, X) -> append_3_out_gaa₃([]_0, X, X)
append_3_in_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_append_3_in_1_gaa₅(X, Xs, Ys, Zs, append_3_in_gaa₃(Xs, Ys, Zs))
if_append_3_in_1_gaa₅(X, Xs, Ys, Zs, append_3_out_gaa₃(Xs, Ys, Zs)) -> append_3_out_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_fl_3_in_1_gaa₅(E, X, R, Z, append_3_out_gaa₃(E, Y, R)) -> if_fl_3_in_2_gaa₆(E, X, R, Z, Y, fl_3_in_gaa₃(X, Y, Z))
if_fl_3_in_2_gaa₆(E, X, R, Z, Y, fl_3_out_gaa₃(X, Y, Z)) -> fl_3_out_gaa₃(._2₂(E, X), R, s_1₁(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_3_in_gaa₃([]_0, []_0, 0_0) -> fl_3_out_gaa₃([]_0, []_0, 0_0)
fl_3_in_gaa₃(._2₂(E, X), R, s_1₁(Z)) -> if_fl_3_in_1_gaa₅(E, X, R, Z, append_3_in_gaa₃(E, Y, R))
append_3_in_gaa₃([]_0, X, X) -> append_3_out_gaa₃([]_0, X, X)
append_3_in_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_append_3_in_1_gaa₅(X, Xs, Ys, Zs, append_3_in_gaa₃(Xs, Ys, Zs))
if_append_3_in_1_gaa₅(X, Xs, Ys, Zs, append_3_out_gaa₃(Xs, Ys, Zs)) -> append_3_out_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_fl_3_in_1_gaa₅(E, X, R, Z, append_3_out_gaa₃(E, Y, R)) -> if_fl_3_in_2_gaa₆(E, X, R, Z, Y, fl_3_in_gaa₃(X, Y, Z))
if_fl_3_in_2_gaa₆(E, X, R, Z, Y, fl_3_out_gaa₃(X, Y, Z)) -> fl_3_out_gaa₃(._2₂(E, X), R, s_1₁(Z))

FL_3_IN_GAA₃(._2₂(E, X), R, s_1₁(Z)) -> IF_FL_3_IN_1_GAA₅(E, X, R, Z, append_3_in_gaa₃(E, Y, R))
FL_3_IN_GAA₃(._2₂(E, X), R, s_1₁(Z)) -> APPEND_3_IN_GAA₃(E, Y, R)
APPEND_3_IN_GAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> IF_APPEND_3_IN_1_GAA₅(X, Xs, Ys, Zs, append_3_in_gaa₃(Xs, Ys, Zs))
APPEND_3_IN_GAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APPEND_3_IN_GAA₃(Xs, Ys, Zs)
IF_FL_3_IN_1_GAA₅(E, X, R, Z, append_3_out_gaa₃(E, Y, R)) -> IF_FL_3_IN_2_GAA₆(E, X, R, Z, Y, fl_3_in_gaa₃(X, Y, Z))
IF_FL_3_IN_1_GAA₅(E, X, R, Z, append_3_out_gaa₃(E, Y, R)) -> FL_3_IN_GAA₃(X, Y, Z)

The TRS R consists of the following rules:

fl_3_in_gaa₃([]_0, []_0, 0_0) -> fl_3_out_gaa₃([]_0, []_0, 0_0)
fl_3_in_gaa₃(._2₂(E, X), R, s_1₁(Z)) -> if_fl_3_in_1_gaa₅(E, X, R, Z, append_3_in_gaa₃(E, Y, R))
append_3_in_gaa₃([]_0, X, X) -> append_3_out_gaa₃([]_0, X, X)
append_3_in_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_append_3_in_1_gaa₅(X, Xs, Ys, Zs, append_3_in_gaa₃(Xs, Ys, Zs))
if_append_3_in_1_gaa₅(X, Xs, Ys, Zs, append_3_out_gaa₃(Xs, Ys, Zs)) -> append_3_out_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_fl_3_in_1_gaa₅(E, X, R, Z, append_3_out_gaa₃(E, Y, R)) -> if_fl_3_in_2_gaa₆(E, X, R, Z, Y, fl_3_in_gaa₃(X, Y, Z))
if_fl_3_in_2_gaa₆(E, X, R, Z, Y, fl_3_out_gaa₃(X, Y, Z)) -> fl_3_out_gaa₃(._2₂(E, X), R, s_1₁(Z))

The argument filtering Pi contains the following mapping:
fl_3_in_gaa₃(x1, x2, x3) = fl_3_in_gaa₁(x1)
[]_0 = []_0
0_0 = 0_0
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
fl_3_out_gaa₃(x1, x2, x3) = fl_3_out_gaa₁(x3)
if_fl_3_in_1_gaa₅(x1, x2, x3, x4, x5) = if_fl_3_in_1_gaa₂(x2, x5)
append_3_in_gaa₃(x1, x2, x3) = append_3_in_gaa₁(x1)
append_3_out_gaa₃(x1, x2, x3) = append_3_out_gaa
if_append_3_in_1_gaa₅(x1, x2, x3, x4, x5) = if_append_3_in_1_gaa₁(x5)
if_fl_3_in_2_gaa₆(x1, x2, x3, x4, x5, x6) = if_fl_3_in_2_gaa₁(x6)
APPEND_3_IN_GAA₃(x1, x2, x3) = APPEND_3_IN_GAA₁(x1)
IF_FL_3_IN_1_GAA₅(x1, x2, x3, x4, x5) = IF_FL_3_IN_1_GAA₂(x2, x5)
IF_FL_3_IN_2_GAA₆(x1, x2, x3, x4, x5, x6) = IF_FL_3_IN_2_GAA₁(x6)
IF_APPEND_3_IN_1_GAA₅(x1, x2, x3, x4, x5) = IF_APPEND_3_IN_1_GAA₁(x5)
FL_3_IN_GAA₃(x1, x2, x3) = FL_3_IN_GAA₁(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:

FL_3_IN_GAA₃(._2₂(E, X), R, s_1₁(Z)) -> IF_FL_3_IN_1_GAA₅(E, X, R, Z, append_3_in_gaa₃(E, Y, R))
FL_3_IN_GAA₃(._2₂(E, X), R, s_1₁(Z)) -> APPEND_3_IN_GAA₃(E, Y, R)
APPEND_3_IN_GAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> IF_APPEND_3_IN_1_GAA₅(X, Xs, Ys, Zs, append_3_in_gaa₃(Xs, Ys, Zs))
APPEND_3_IN_GAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APPEND_3_IN_GAA₃(Xs, Ys, Zs)
IF_FL_3_IN_1_GAA₅(E, X, R, Z, append_3_out_gaa₃(E, Y, R)) -> IF_FL_3_IN_2_GAA₆(E, X, R, Z, Y, fl_3_in_gaa₃(X, Y, Z))
IF_FL_3_IN_1_GAA₅(E, X, R, Z, append_3_out_gaa₃(E, Y, R)) -> FL_3_IN_GAA₃(X, Y, Z)

The TRS R consists of the following rules:

fl_3_in_gaa₃([]_0, []_0, 0_0) -> fl_3_out_gaa₃([]_0, []_0, 0_0)
fl_3_in_gaa₃(._2₂(E, X), R, s_1₁(Z)) -> if_fl_3_in_1_gaa₅(E, X, R, Z, append_3_in_gaa₃(E, Y, R))
append_3_in_gaa₃([]_0, X, X) -> append_3_out_gaa₃([]_0, X, X)
append_3_in_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_append_3_in_1_gaa₅(X, Xs, Ys, Zs, append_3_in_gaa₃(Xs, Ys, Zs))
if_append_3_in_1_gaa₅(X, Xs, Ys, Zs, append_3_out_gaa₃(Xs, Ys, Zs)) -> append_3_out_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_fl_3_in_1_gaa₅(E, X, R, Z, append_3_out_gaa₃(E, Y, R)) -> if_fl_3_in_2_gaa₆(E, X, R, Z, Y, fl_3_in_gaa₃(X, Y, Z))
if_fl_3_in_2_gaa₆(E, X, R, Z, Y, fl_3_out_gaa₃(X, Y, Z)) -> fl_3_out_gaa₃(._2₂(E, X), R, s_1₁(Z))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

APPEND_3_IN_GAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APPEND_3_IN_GAA₃(Xs, Ys, Zs)

The TRS R consists of the following rules:

fl_3_in_gaa₃([]_0, []_0, 0_0) -> fl_3_out_gaa₃([]_0, []_0, 0_0)
fl_3_in_gaa₃(._2₂(E, X), R, s_1₁(Z)) -> if_fl_3_in_1_gaa₅(E, X, R, Z, append_3_in_gaa₃(E, Y, R))
append_3_in_gaa₃([]_0, X, X) -> append_3_out_gaa₃([]_0, X, X)
append_3_in_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_append_3_in_1_gaa₅(X, Xs, Ys, Zs, append_3_in_gaa₃(Xs, Ys, Zs))
if_append_3_in_1_gaa₅(X, Xs, Ys, Zs, append_3_out_gaa₃(Xs, Ys, Zs)) -> append_3_out_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_fl_3_in_1_gaa₅(E, X, R, Z, append_3_out_gaa₃(E, Y, R)) -> if_fl_3_in_2_gaa₆(E, X, R, Z, Y, fl_3_in_gaa₃(X, Y, Z))
if_fl_3_in_2_gaa₆(E, X, R, Z, Y, fl_3_out_gaa₃(X, Y, Z)) -> fl_3_out_gaa₃(._2₂(E, X), R, s_1₁(Z))

The argument filtering Pi contains the following mapping:
fl_3_in_gaa₃(x1, x2, x3) = fl_3_in_gaa₁(x1)
[]_0 = []_0
0_0 = 0_0
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
fl_3_out_gaa₃(x1, x2, x3) = fl_3_out_gaa₁(x3)
if_fl_3_in_1_gaa₅(x1, x2, x3, x4, x5) = if_fl_3_in_1_gaa₂(x2, x5)
append_3_in_gaa₃(x1, x2, x3) = append_3_in_gaa₁(x1)
append_3_out_gaa₃(x1, x2, x3) = append_3_out_gaa
if_append_3_in_1_gaa₅(x1, x2, x3, x4, x5) = if_append_3_in_1_gaa₁(x5)
if_fl_3_in_2_gaa₆(x1, x2, x3, x4, x5, x6) = if_fl_3_in_2_gaa₁(x6)
APPEND_3_IN_GAA₃(x1, x2, x3) = APPEND_3_IN_GAA₁(x1)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting 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_3_IN_GAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APPEND_3_IN_GAA₃(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
APPEND_3_IN_GAA₃(x1, x2, x3) = APPEND_3_IN_GAA₁(x1)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem 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_3_IN_GAA₁(._2₂(X, Xs)) -> APPEND_3_IN_GAA₁(Xs)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {APPEND_3_IN_GAA₁}.
By using the subterm criterion together with the size-change analysis 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_3_IN_GAA₁(._2₂(X, Xs)) -> APPEND_3_IN_GAA₁(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_3_IN_GAA₃(._2₂(E, X), R, s_1₁(Z)) -> IF_FL_3_IN_1_GAA₅(E, X, R, Z, append_3_in_gaa₃(E, Y, R))
IF_FL_3_IN_1_GAA₅(E, X, R, Z, append_3_out_gaa₃(E, Y, R)) -> FL_3_IN_GAA₃(X, Y, Z)

The TRS R consists of the following rules:

fl_3_in_gaa₃([]_0, []_0, 0_0) -> fl_3_out_gaa₃([]_0, []_0, 0_0)
fl_3_in_gaa₃(._2₂(E, X), R, s_1₁(Z)) -> if_fl_3_in_1_gaa₅(E, X, R, Z, append_3_in_gaa₃(E, Y, R))
append_3_in_gaa₃([]_0, X, X) -> append_3_out_gaa₃([]_0, X, X)
append_3_in_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_append_3_in_1_gaa₅(X, Xs, Ys, Zs, append_3_in_gaa₃(Xs, Ys, Zs))
if_append_3_in_1_gaa₅(X, Xs, Ys, Zs, append_3_out_gaa₃(Xs, Ys, Zs)) -> append_3_out_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_fl_3_in_1_gaa₅(E, X, R, Z, append_3_out_gaa₃(E, Y, R)) -> if_fl_3_in_2_gaa₆(E, X, R, Z, Y, fl_3_in_gaa₃(X, Y, Z))
if_fl_3_in_2_gaa₆(E, X, R, Z, Y, fl_3_out_gaa₃(X, Y, Z)) -> fl_3_out_gaa₃(._2₂(E, X), R, s_1₁(Z))

The argument filtering Pi contains the following mapping:
fl_3_in_gaa₃(x1, x2, x3) = fl_3_in_gaa₁(x1)
[]_0 = []_0
0_0 = 0_0
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
fl_3_out_gaa₃(x1, x2, x3) = fl_3_out_gaa₁(x3)
if_fl_3_in_1_gaa₅(x1, x2, x3, x4, x5) = if_fl_3_in_1_gaa₂(x2, x5)
append_3_in_gaa₃(x1, x2, x3) = append_3_in_gaa₁(x1)
append_3_out_gaa₃(x1, x2, x3) = append_3_out_gaa
if_append_3_in_1_gaa₅(x1, x2, x3, x4, x5) = if_append_3_in_1_gaa₁(x5)
if_fl_3_in_2_gaa₆(x1, x2, x3, x4, x5, x6) = if_fl_3_in_2_gaa₁(x6)
IF_FL_3_IN_1_GAA₅(x1, x2, x3, x4, x5) = IF_FL_3_IN_1_GAA₂(x2, x5)
FL_3_IN_GAA₃(x1, x2, x3) = FL_3_IN_GAA₁(x1)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting 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_3_IN_GAA₃(._2₂(E, X), R, s_1₁(Z)) -> IF_FL_3_IN_1_GAA₅(E, X, R, Z, append_3_in_gaa₃(E, Y, R))
IF_FL_3_IN_1_GAA₅(E, X, R, Z, append_3_out_gaa₃(E, Y, R)) -> FL_3_IN_GAA₃(X, Y, Z)

The TRS R consists of the following rules:

append_3_in_gaa₃([]_0, X, X) -> append_3_out_gaa₃([]_0, X, X)
append_3_in_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_append_3_in_1_gaa₅(X, Xs, Ys, Zs, append_3_in_gaa₃(Xs, Ys, Zs))
if_append_3_in_1_gaa₅(X, Xs, Ys, Zs, append_3_out_gaa₃(Xs, Ys, Zs)) -> append_3_out_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))

The argument filtering Pi contains the following mapping:
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
append_3_in_gaa₃(x1, x2, x3) = append_3_in_gaa₁(x1)
append_3_out_gaa₃(x1, x2, x3) = append_3_out_gaa
if_append_3_in_1_gaa₅(x1, x2, x3, x4, x5) = if_append_3_in_1_gaa₁(x5)
IF_FL_3_IN_1_GAA₅(x1, x2, x3, x4, x5) = IF_FL_3_IN_1_GAA₂(x2, x5)
FL_3_IN_GAA₃(x1, x2, x3) = FL_3_IN_GAA₁(x1)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem 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_3_IN_GAA₁(._2₂(E, X)) -> IF_FL_3_IN_1_GAA₂(X, append_3_in_gaa₁(E))
IF_FL_3_IN_1_GAA₂(X, append_3_out_gaa) -> FL_3_IN_GAA₁(X)

The TRS R consists of the following rules:

append_3_in_gaa₁([]_0) -> append_3_out_gaa
append_3_in_gaa₁(._2₂(X, Xs)) -> if_append_3_in_1_gaa₁(append_3_in_gaa₁(Xs))
if_append_3_in_1_gaa₁(append_3_out_gaa) -> append_3_out_gaa

The set Q consists of the following terms:

append_3_in_gaa₁(x0)
if_append_3_in_1_gaa₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_FL_3_IN_1_GAA₂, FL_3_IN_GAA₁}.
By using the subterm criterion together with the size-change analysis 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:

IF_FL_3_IN_1_GAA₂(X, append_3_out_gaa) -> FL_3_IN_GAA₁(X)
The graph contains the following edges 1 >= 1
FL_3_IN_GAA₁(._2₂(E, X)) -> IF_FL_3_IN_1_GAA₂(X, append_3_in_gaa₁(E))
The graph contains the following edges 1 > 1