Left Termination proof of /tmp/tmp3jTOAB/flatlength-fbf.pl

Left Termination of the query pattern fl_in_3(a, g, a) w.r.t. the given Prolog program could not be shown:

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ 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(a,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

  ↳ PrologToPiTRSProof

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(x2)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x5)
append_in(x1, x2, x3) = append_in(x3)
U3(x1, x2, x3, x4, x5) = U3(x1, x5)
[] = []
append_out(x1, x2, x3) = append_out(x1, x2)
U2(x1, x2, x3, x4, x5) = U2(x1, x5)
0 = 0
fl_out(x1, x2, x3) = fl_out(x1, 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(x2)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x5)
append_in(x1, x2, x3) = append_in(x3)
U3(x1, x2, x3, x4, x5) = U3(x1, x5)
[] = []
append_out(x1, x2, x3) = append_out(x1, x2)
U2(x1, x2, x3, x4, x5) = U2(x1, x5)
0 = 0
fl_out(x1, x2, x3) = fl_out(x1, x3)
U3¹(x1, x2, x3, x4, x5) = U3¹(x1, x5)
APPEND_IN(x1, x2, x3) = APPEND_IN(x3)
FL_IN(x1, x2, x3) = FL_IN(x2)
U2¹(x1, x2, x3, x4, x5) = U2¹(x1, x5)
U1¹(x1, x2, x3, x4, x5) = U1¹(x5)

We have to consider all (P,R,Pi)-chains

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

  ↳ PrologToPiTRSProof

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(x2)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x5)
append_in(x1, x2, x3) = append_in(x3)
U3(x1, x2, x3, x4, x5) = U3(x1, x5)
[] = []
append_out(x1, x2, x3) = append_out(x1, x2)
U2(x1, x2, x3, x4, x5) = U2(x1, x5)
0 = 0
fl_out(x1, x2, x3) = fl_out(x1, x3)
U3¹(x1, x2, x3, x4, x5) = U3¹(x1, x5)
APPEND_IN(x1, x2, x3) = APPEND_IN(x3)
FL_IN(x1, x2, x3) = FL_IN(x2)
U2¹(x1, x2, x3, x4, x5) = U2¹(x1, x5)
U1¹(x1, x2, x3, x4, x5) = U1¹(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

  ↳ PrologToPiTRSProof

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(x2)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x5)
append_in(x1, x2, x3) = append_in(x3)
U3(x1, x2, x3, x4, x5) = U3(x1, x5)
[] = []
append_out(x1, x2, x3) = append_out(x1, x2)
U2(x1, x2, x3, x4, x5) = U2(x1, x5)
0 = 0
fl_out(x1, x2, x3) = fl_out(x1, x3)
APPEND_IN(x1, x2, x3) = APPEND_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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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(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

  ↳ PrologToPiTRSProof

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

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

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

  ↳ PrologToPiTRSProof

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(x2)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x5)
append_in(x1, x2, x3) = append_in(x3)
U3(x1, x2, x3, x4, x5) = U3(x1, x5)
[] = []
append_out(x1, x2, x3) = append_out(x1, x2)
U2(x1, x2, x3, x4, x5) = U2(x1, x5)
0 = 0
fl_out(x1, x2, x3) = fl_out(x1, x3)
FL_IN(x1, x2, x3) = FL_IN(x2)
U1¹(x1, x2, x3, x4, x5) = U1¹(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

  ↳ PrologToPiTRSProof

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(x3)
U3(x1, x2, x3, x4, x5) = U3(x1, x5)
[] = []
append_out(x1, x2, x3) = append_out(x1, x2)
FL_IN(x1, x2, x3) = FL_IN(x2)
U1¹(x1, x2, x3, x4, x5) = U1¹(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

                        ↳ RuleRemovalProof

  ↳ PrologToPiTRSProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

append_in(x0)
U3(x0, x1)

We have to consider all (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

append_in(.(X, Zs)) → U3(X, append_in(Zs))

Used ordering: POLO with Polynomial interpretation [25]:

POL(.(x₁, x₂)) = 1 + x₁ + x₂
POL(FL_IN(x₁)) = 2·x₁
POL(U1¹(x₁)) = x₁
POL(U3(x₁, x₂)) = 1 + x₁ + x₂
POL([]) = 0
POL(append_in(x₁)) = 2·x₁
POL(append_out(x₁, x₂)) = x₁ + 2·x₂

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ UsableRulesProof

  ↳ PrologToPiTRSProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

append_in(x0)
U3(x0, x1)

We have to consider all (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QReductionProof

  ↳ PrologToPiTRSProof

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

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

The TRS R consists of the following rules:

append_in(X) → append_out([], X)

The set Q consists of the following terms:

append_in(x0)
U3(x0, x1)

We have to consider all (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

U3(x0, x1)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QReductionProof

                                  ↳ QDP

                                    ↳ Rewriting

  ↳ PrologToPiTRSProof

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

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

The TRS R consists of the following rules:

append_in(X) → append_out([], X)

The set Q consists of the following terms:

append_in(x0)

We have to consider all (P,Q,R)-chains.
By rewriting [15] the rule FL_IN(R) → U1¹(append_in(R)) at position [0] we obtained the following new rules:

FL_IN(R) → U1¹(append_out([], R))

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QReductionProof

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ UsableRulesProof

  ↳ PrologToPiTRSProof

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

FL_IN(R) → U1¹(append_out([], R))
U1¹(append_out(E, Y)) → FL_IN(Y)

The TRS R consists of the following rules:

append_in(X) → append_out([], X)

The set Q consists of the following terms:

append_in(x0)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QReductionProof

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ QReductionProof

  ↳ PrologToPiTRSProof

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

FL_IN(R) → U1¹(append_out([], R))
U1¹(append_out(E, Y)) → FL_IN(Y)

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

append_in(x0)

We have to consider all (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

append_in(x0)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QReductionProof

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ QReductionProof

                                              ↳ QDP

                                                ↳ Instantiation

  ↳ PrologToPiTRSProof

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

FL_IN(R) → U1¹(append_out([], R))
U1¹(append_out(E, Y)) → FL_IN(Y)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule U1¹(append_out(E, Y)) → FL_IN(Y) we obtained the following new rules:

U1¹(append_out([], z0)) → FL_IN(z0)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QReductionProof

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ QReductionProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ NonTerminationProof

  ↳ PrologToPiTRSProof

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

U1¹(append_out([], z0)) → FL_IN(z0)
FL_IN(R) → U1¹(append_out([], R))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

U1¹(append_out([], z0)) → FL_IN(z0)
FL_IN(R) → U1¹(append_out([], R))

The TRS R consists of the following rules:none

s = FL_IN(R) evaluates to t =FL_IN(R)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

FL_IN(R) → U1¹(append_out([], R))
with rule FL_IN(R') → U1¹(append_out([], R')) at position [] and matcher [R' / R]

U1¹(append_out([], R)) → FL_IN(R)
with rule U1¹(append_out([], z0)) → FL_IN(z0)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence

All these steps are and every following step will be a correct step w.r.t to Q.

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

  ↳ 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(x2)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x3, x5)
append_in(x1, x2, x3) = append_in(x3)
U3(x1, x2, x3, x4, x5) = U3(x1, x4, x5)
[] = []
append_out(x1, x2, x3) = append_out(x1, x2, x3)
U2(x1, x2, x3, x4, x5) = U2(x1, x3, x5)
0 = 0
fl_out(x1, x2, x3) = fl_out(x1, x2, 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(x2)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x3, x5)
append_in(x1, x2, x3) = append_in(x3)
U3(x1, x2, x3, x4, x5) = U3(x1, x4, x5)
[] = []
append_out(x1, x2, x3) = append_out(x1, x2, x3)
U2(x1, x2, x3, x4, x5) = U2(x1, x3, x5)
0 = 0
fl_out(x1, x2, x3) = fl_out(x1, x2, x3)
U3¹(x1, x2, x3, x4, x5) = U3¹(x1, x4, x5)
APPEND_IN(x1, x2, x3) = APPEND_IN(x3)
FL_IN(x1, x2, x3) = FL_IN(x2)
U2¹(x1, x2, x3, x4, x5) = U2¹(x1, x3, x5)
U1¹(x1, x2, x3, x4, x5) = U1¹(x3, x5)

We have to consider all (P,R,Pi)-chains

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ 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(x2)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x3, x5)
append_in(x1, x2, x3) = append_in(x3)
U3(x1, x2, x3, x4, x5) = U3(x1, x4, x5)
[] = []
append_out(x1, x2, x3) = append_out(x1, x2, x3)
U2(x1, x2, x3, x4, x5) = U2(x1, x3, x5)
0 = 0
fl_out(x1, x2, x3) = fl_out(x1, x2, x3)
U3¹(x1, x2, x3, x4, x5) = U3¹(x1, x4, x5)
APPEND_IN(x1, x2, x3) = APPEND_IN(x3)
FL_IN(x1, x2, x3) = FL_IN(x2)
U2¹(x1, x2, x3, x4, x5) = U2¹(x1, x3, x5)
U1¹(x1, x2, x3, x4, x5) = U1¹(x3, 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

  ↳ 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(x2)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x3, x5)
append_in(x1, x2, x3) = append_in(x3)
U3(x1, x2, x3, x4, x5) = U3(x1, x4, x5)
[] = []
append_out(x1, x2, x3) = append_out(x1, x2, x3)
U2(x1, x2, x3, x4, x5) = U2(x1, x3, x5)
0 = 0
fl_out(x1, x2, x3) = fl_out(x1, x2, x3)
APPEND_IN(x1, x2, x3) = APPEND_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

  ↳ 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)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ 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, Zs)) → APPEND_IN(Zs)

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

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

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ 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(x2)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x3, x5)
append_in(x1, x2, x3) = append_in(x3)
U3(x1, x2, x3, x4, x5) = U3(x1, x4, x5)
[] = []
append_out(x1, x2, x3) = append_out(x1, x2, x3)
U2(x1, x2, x3, x4, x5) = U2(x1, x3, x5)
0 = 0
fl_out(x1, x2, x3) = fl_out(x1, x2, x3)
FL_IN(x1, x2, x3) = FL_IN(x2)
U1¹(x1, x2, x3, x4, x5) = U1¹(x3, 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

  ↳ 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(x3)
U3(x1, x2, x3, x4, x5) = U3(x1, x4, x5)
[] = []
append_out(x1, x2, x3) = append_out(x1, x2, x3)
FL_IN(x1, x2, x3) = FL_IN(x2)
U1¹(x1, x2, x3, x4, x5) = U1¹(x3, 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

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

append_in(x0)
U3(x0, x1, x2)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule FL_IN(R) → U1¹(R, append_in(R)) at position [1] we obtained the following new rules:

FL_IN(x0) → U1¹(x0, append_out([], x0, x0))
FL_IN(.(x0, x1)) → U1¹(.(x0, x1), U3(x0, x1, append_in(x1)))

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Instantiation

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

FL_IN(x0) → U1¹(x0, append_out([], x0, x0))
U1¹(R, append_out(E, Y, R)) → FL_IN(Y)
FL_IN(.(x0, x1)) → U1¹(.(x0, x1), U3(x0, x1, append_in(x1)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

append_in(x0)
U3(x0, x1, x2)

We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule U1¹(R, append_out(E, Y, R)) → FL_IN(Y) we obtained the following new rules:

U1¹(z0, append_out([], z0, z0)) → FL_IN(z0)
U1¹(.(z0, z1), append_out(x1, x2, .(z0, z1))) → FL_IN(x2)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ QDPOrderProof

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

FL_IN(x0) → U1¹(x0, append_out([], x0, x0))
U1¹(z0, append_out([], z0, z0)) → FL_IN(z0)
U1¹(.(z0, z1), append_out(x1, x2, .(z0, z1))) → FL_IN(x2)
FL_IN(.(x0, x1)) → U1¹(.(x0, x1), U3(x0, x1, append_in(x1)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

append_in(x0)
U3(x0, x1, x2)

We have to consider all (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

FL_IN(.(x0, x1)) → U1¹(.(x0, x1), U3(x0, x1, append_in(x1)))

The remaining pairs can at least be oriented weakly.

FL_IN(x0) → U1¹(x0, append_out([], x0, x0))
U1¹(z0, append_out([], z0, z0)) → FL_IN(z0)
U1¹(.(z0, z1), append_out(x1, x2, .(z0, z1))) → FL_IN(x2)

Used ordering: Polynomial interpretation [25]:

POL(.(x₁, x₂)) = 1 + x₁ + x₂
POL(FL_IN(x₁)) = 1 + x₁
POL(U1¹(x₁, x₂)) = 1 + x₂
POL(U3(x₁, x₂, x₃)) = x₁ + x₃
POL([]) = 0
POL(append_in(x₁)) = x₁
POL(append_out(x₁, x₂, x₃)) = x₂

The following usable rules [17] were oriented:

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

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ UsableRulesProof

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

FL_IN(x0) → U1¹(x0, append_out([], x0, x0))
U1¹(z0, append_out([], z0, z0)) → FL_IN(z0)
U1¹(.(z0, z1), append_out(x1, x2, .(z0, z1))) → FL_IN(x2)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

append_in(x0)
U3(x0, x1, x2)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

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

FL_IN(x0) → U1¹(x0, append_out([], x0, x0))
U1¹(z0, append_out([], z0, z0)) → FL_IN(z0)
U1¹(.(z0, z1), append_out(x1, x2, .(z0, z1))) → FL_IN(x2)

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

append_in(x0)
U3(x0, x1, x2)

We have to consider all (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

append_in(x0)
U3(x0, x1, x2)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

                                          ↳ QDP

                                            ↳ Instantiation

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

FL_IN(x0) → U1¹(x0, append_out([], x0, x0))
U1¹(z0, append_out([], z0, z0)) → FL_IN(z0)
U1¹(.(z0, z1), append_out(x1, x2, .(z0, z1))) → FL_IN(x2)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule U1¹(.(z0, z1), append_out(x1, x2, .(z0, z1))) → FL_IN(x2) we obtained the following new rules:

U1¹(.(x0, x1), append_out([], .(x0, x1), .(x0, x1))) → FL_IN(.(x0, x1))

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

                                          ↳ QDP

                                            ↳ Instantiation

                                              ↳ QDP

                                                ↳ NonTerminationProof

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

FL_IN(x0) → U1¹(x0, append_out([], x0, x0))
U1¹(z0, append_out([], z0, z0)) → FL_IN(z0)
U1¹(.(x0, x1), append_out([], .(x0, x1), .(x0, x1))) → FL_IN(.(x0, x1))

The TRS R consists of the following rules:none

s = U1¹(z0, append_out([], z0, z0)) evaluates to t =U1¹(z0, append_out([], z0, z0))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

U1¹(z0, append_out([], z0, z0)) → FL_IN(z0)
with rule U1¹(z0', append_out([], z0', z0')) → FL_IN(z0') at position [] and matcher [z0' / z0]

FL_IN(z0) → U1¹(z0, append_out([], z0, z0))
with rule FL_IN(x0) → U1¹(x0, append_out([], x0, x0))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence

All these steps are and every following step will be a correct step w.r.t to Q.