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

Left Termination of the query pattern fl(f,b,f) w.r.t. the given Prolog program could not be shown:

↳ PROLOG

  ↳ PrologToPiTRSProof

  ↳ 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: (f,b,f)
append₃: (f,f,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

fl_3_in_aga₃([]_0, []_0, 0_0) -> fl_3_out_aga₃([]_0, []_0, 0_0)
fl_3_in_aga₃(._2₂(E, X), R, s_1₁(Z)) -> if_fl_3_in_1_aga₅(E, X, R, Z, append_3_in_aag₃(E, Y, R))
append_3_in_aag₃([]_0, X, X) -> append_3_out_aag₃([]_0, X, X)
append_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_in_aag₃(Xs, Ys, Zs))
if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_out_aag₃(Xs, Ys, Zs)) -> append_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_fl_3_in_1_aga₅(E, X, R, Z, append_3_out_aag₃(E, Y, R)) -> if_fl_3_in_2_aga₆(E, X, R, Z, Y, fl_3_in_aga₃(X, Y, Z))
if_fl_3_in_2_aga₆(E, X, R, Z, Y, fl_3_out_aga₃(X, Y, Z)) -> fl_3_out_aga₃(._2₂(E, X), R, s_1₁(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_3_in_aga₃([]_0, []_0, 0_0) -> fl_3_out_aga₃([]_0, []_0, 0_0)
fl_3_in_aga₃(._2₂(E, X), R, s_1₁(Z)) -> if_fl_3_in_1_aga₅(E, X, R, Z, append_3_in_aag₃(E, Y, R))
append_3_in_aag₃([]_0, X, X) -> append_3_out_aag₃([]_0, X, X)
append_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_in_aag₃(Xs, Ys, Zs))
if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_out_aag₃(Xs, Ys, Zs)) -> append_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_fl_3_in_1_aga₅(E, X, R, Z, append_3_out_aag₃(E, Y, R)) -> if_fl_3_in_2_aga₆(E, X, R, Z, Y, fl_3_in_aga₃(X, Y, Z))
if_fl_3_in_2_aga₆(E, X, R, Z, Y, fl_3_out_aga₃(X, Y, Z)) -> fl_3_out_aga₃(._2₂(E, X), R, s_1₁(Z))

FL_3_IN_AGA₃(._2₂(E, X), R, s_1₁(Z)) -> IF_FL_3_IN_1_AGA₅(E, X, R, Z, append_3_in_aag₃(E, Y, R))
FL_3_IN_AGA₃(._2₂(E, X), R, s_1₁(Z)) -> APPEND_3_IN_AAG₃(E, Y, R)
APPEND_3_IN_AAG₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> IF_APPEND_3_IN_1_AAG₅(X, Xs, Ys, Zs, append_3_in_aag₃(Xs, Ys, Zs))
APPEND_3_IN_AAG₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APPEND_3_IN_AAG₃(Xs, Ys, Zs)
IF_FL_3_IN_1_AGA₅(E, X, R, Z, append_3_out_aag₃(E, Y, R)) -> IF_FL_3_IN_2_AGA₆(E, X, R, Z, Y, fl_3_in_aga₃(X, Y, Z))
IF_FL_3_IN_1_AGA₅(E, X, R, Z, append_3_out_aag₃(E, Y, R)) -> FL_3_IN_AGA₃(X, Y, Z)

The TRS R consists of the following rules:

fl_3_in_aga₃([]_0, []_0, 0_0) -> fl_3_out_aga₃([]_0, []_0, 0_0)
fl_3_in_aga₃(._2₂(E, X), R, s_1₁(Z)) -> if_fl_3_in_1_aga₅(E, X, R, Z, append_3_in_aag₃(E, Y, R))
append_3_in_aag₃([]_0, X, X) -> append_3_out_aag₃([]_0, X, X)
append_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_in_aag₃(Xs, Ys, Zs))
if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_out_aag₃(Xs, Ys, Zs)) -> append_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_fl_3_in_1_aga₅(E, X, R, Z, append_3_out_aag₃(E, Y, R)) -> if_fl_3_in_2_aga₆(E, X, R, Z, Y, fl_3_in_aga₃(X, Y, Z))
if_fl_3_in_2_aga₆(E, X, R, Z, Y, fl_3_out_aga₃(X, Y, Z)) -> fl_3_out_aga₃(._2₂(E, X), R, s_1₁(Z))

The argument filtering Pi contains the following mapping:
fl_3_in_aga₃(x1, x2, x3) = fl_3_in_aga₁(x2)
[]_0 = []_0
0_0 = 0_0
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
fl_3_out_aga₃(x1, x2, x3) = fl_3_out_aga₂(x1, x3)
if_fl_3_in_1_aga₅(x1, x2, x3, x4, x5) = if_fl_3_in_1_aga₁(x5)
append_3_in_aag₃(x1, x2, x3) = append_3_in_aag₁(x3)
append_3_out_aag₃(x1, x2, x3) = append_3_out_aag₂(x1, x2)
if_append_3_in_1_aag₅(x1, x2, x3, x4, x5) = if_append_3_in_1_aag₂(x1, x5)
if_fl_3_in_2_aga₆(x1, x2, x3, x4, x5, x6) = if_fl_3_in_2_aga₂(x1, x6)
IF_APPEND_3_IN_1_AAG₅(x1, x2, x3, x4, x5) = IF_APPEND_3_IN_1_AAG₂(x1, x5)
FL_3_IN_AGA₃(x1, x2, x3) = FL_3_IN_AGA₁(x2)
APPEND_3_IN_AAG₃(x1, x2, x3) = APPEND_3_IN_AAG₁(x3)
IF_FL_3_IN_2_AGA₆(x1, x2, x3, x4, x5, x6) = IF_FL_3_IN_2_AGA₂(x1, x6)
IF_FL_3_IN_1_AGA₅(x1, x2, x3, x4, x5) = IF_FL_3_IN_1_AGA₁(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_3_IN_AGA₃(._2₂(E, X), R, s_1₁(Z)) -> IF_FL_3_IN_1_AGA₅(E, X, R, Z, append_3_in_aag₃(E, Y, R))
FL_3_IN_AGA₃(._2₂(E, X), R, s_1₁(Z)) -> APPEND_3_IN_AAG₃(E, Y, R)
APPEND_3_IN_AAG₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> IF_APPEND_3_IN_1_AAG₅(X, Xs, Ys, Zs, append_3_in_aag₃(Xs, Ys, Zs))
APPEND_3_IN_AAG₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APPEND_3_IN_AAG₃(Xs, Ys, Zs)
IF_FL_3_IN_1_AGA₅(E, X, R, Z, append_3_out_aag₃(E, Y, R)) -> IF_FL_3_IN_2_AGA₆(E, X, R, Z, Y, fl_3_in_aga₃(X, Y, Z))
IF_FL_3_IN_1_AGA₅(E, X, R, Z, append_3_out_aag₃(E, Y, R)) -> FL_3_IN_AGA₃(X, Y, Z)

The TRS R consists of the following rules:

fl_3_in_aga₃([]_0, []_0, 0_0) -> fl_3_out_aga₃([]_0, []_0, 0_0)
fl_3_in_aga₃(._2₂(E, X), R, s_1₁(Z)) -> if_fl_3_in_1_aga₅(E, X, R, Z, append_3_in_aag₃(E, Y, R))
append_3_in_aag₃([]_0, X, X) -> append_3_out_aag₃([]_0, X, X)
append_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_in_aag₃(Xs, Ys, Zs))
if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_out_aag₃(Xs, Ys, Zs)) -> append_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_fl_3_in_1_aga₅(E, X, R, Z, append_3_out_aag₃(E, Y, R)) -> if_fl_3_in_2_aga₆(E, X, R, Z, Y, fl_3_in_aga₃(X, Y, Z))
if_fl_3_in_2_aga₆(E, X, R, Z, Y, fl_3_out_aga₃(X, Y, Z)) -> fl_3_out_aga₃(._2₂(E, X), R, s_1₁(Z))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

APPEND_3_IN_AAG₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APPEND_3_IN_AAG₃(Xs, Ys, Zs)

The TRS R consists of the following rules:

fl_3_in_aga₃([]_0, []_0, 0_0) -> fl_3_out_aga₃([]_0, []_0, 0_0)
fl_3_in_aga₃(._2₂(E, X), R, s_1₁(Z)) -> if_fl_3_in_1_aga₅(E, X, R, Z, append_3_in_aag₃(E, Y, R))
append_3_in_aag₃([]_0, X, X) -> append_3_out_aag₃([]_0, X, X)
append_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_in_aag₃(Xs, Ys, Zs))
if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_out_aag₃(Xs, Ys, Zs)) -> append_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_fl_3_in_1_aga₅(E, X, R, Z, append_3_out_aag₃(E, Y, R)) -> if_fl_3_in_2_aga₆(E, X, R, Z, Y, fl_3_in_aga₃(X, Y, Z))
if_fl_3_in_2_aga₆(E, X, R, Z, Y, fl_3_out_aga₃(X, Y, Z)) -> fl_3_out_aga₃(._2₂(E, X), R, s_1₁(Z))

The argument filtering Pi contains the following mapping:
fl_3_in_aga₃(x1, x2, x3) = fl_3_in_aga₁(x2)
[]_0 = []_0
0_0 = 0_0
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
fl_3_out_aga₃(x1, x2, x3) = fl_3_out_aga₂(x1, x3)
if_fl_3_in_1_aga₅(x1, x2, x3, x4, x5) = if_fl_3_in_1_aga₁(x5)
append_3_in_aag₃(x1, x2, x3) = append_3_in_aag₁(x3)
append_3_out_aag₃(x1, x2, x3) = append_3_out_aag₂(x1, x2)
if_append_3_in_1_aag₅(x1, x2, x3, x4, x5) = if_append_3_in_1_aag₂(x1, x5)
if_fl_3_in_2_aga₆(x1, x2, x3, x4, x5, x6) = if_fl_3_in_2_aga₂(x1, x6)
APPEND_3_IN_AAG₃(x1, x2, x3) = APPEND_3_IN_AAG₁(x3)

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

  ↳ PrologToPiTRSProof

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

APPEND_3_IN_AAG₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APPEND_3_IN_AAG₃(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
APPEND_3_IN_AAG₃(x1, x2, x3) = APPEND_3_IN_AAG₁(x3)

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

  ↳ PrologToPiTRSProof

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

APPEND_3_IN_AAG₁(._2₂(X, Zs)) -> APPEND_3_IN_AAG₁(Zs)

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_AAG₁}.
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_AAG₁(._2₂(X, Zs)) -> APPEND_3_IN_AAG₁(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:

IF_FL_3_IN_1_AGA₅(E, X, R, Z, append_3_out_aag₃(E, Y, R)) -> FL_3_IN_AGA₃(X, Y, Z)
FL_3_IN_AGA₃(._2₂(E, X), R, s_1₁(Z)) -> IF_FL_3_IN_1_AGA₅(E, X, R, Z, append_3_in_aag₃(E, Y, R))

The TRS R consists of the following rules:

fl_3_in_aga₃([]_0, []_0, 0_0) -> fl_3_out_aga₃([]_0, []_0, 0_0)
fl_3_in_aga₃(._2₂(E, X), R, s_1₁(Z)) -> if_fl_3_in_1_aga₅(E, X, R, Z, append_3_in_aag₃(E, Y, R))
append_3_in_aag₃([]_0, X, X) -> append_3_out_aag₃([]_0, X, X)
append_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_in_aag₃(Xs, Ys, Zs))
if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_out_aag₃(Xs, Ys, Zs)) -> append_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_fl_3_in_1_aga₅(E, X, R, Z, append_3_out_aag₃(E, Y, R)) -> if_fl_3_in_2_aga₆(E, X, R, Z, Y, fl_3_in_aga₃(X, Y, Z))
if_fl_3_in_2_aga₆(E, X, R, Z, Y, fl_3_out_aga₃(X, Y, Z)) -> fl_3_out_aga₃(._2₂(E, X), R, s_1₁(Z))

The argument filtering Pi contains the following mapping:
fl_3_in_aga₃(x1, x2, x3) = fl_3_in_aga₁(x2)
[]_0 = []_0
0_0 = 0_0
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
fl_3_out_aga₃(x1, x2, x3) = fl_3_out_aga₂(x1, x3)
if_fl_3_in_1_aga₅(x1, x2, x3, x4, x5) = if_fl_3_in_1_aga₁(x5)
append_3_in_aag₃(x1, x2, x3) = append_3_in_aag₁(x3)
append_3_out_aag₃(x1, x2, x3) = append_3_out_aag₂(x1, x2)
if_append_3_in_1_aag₅(x1, x2, x3, x4, x5) = if_append_3_in_1_aag₂(x1, x5)
if_fl_3_in_2_aga₆(x1, x2, x3, x4, x5, x6) = if_fl_3_in_2_aga₂(x1, x6)
FL_3_IN_AGA₃(x1, x2, x3) = FL_3_IN_AGA₁(x2)
IF_FL_3_IN_1_AGA₅(x1, x2, x3, x4, x5) = IF_FL_3_IN_1_AGA₁(x5)

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

  ↳ PrologToPiTRSProof

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

IF_FL_3_IN_1_AGA₅(E, X, R, Z, append_3_out_aag₃(E, Y, R)) -> FL_3_IN_AGA₃(X, Y, Z)
FL_3_IN_AGA₃(._2₂(E, X), R, s_1₁(Z)) -> IF_FL_3_IN_1_AGA₅(E, X, R, Z, append_3_in_aag₃(E, Y, R))

The TRS R consists of the following rules:

append_3_in_aag₃([]_0, X, X) -> append_3_out_aag₃([]_0, X, X)
append_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_in_aag₃(Xs, Ys, Zs))
if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_out_aag₃(Xs, Ys, Zs)) -> append_3_out_aag₃(._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_aag₃(x1, x2, x3) = append_3_in_aag₁(x3)
append_3_out_aag₃(x1, x2, x3) = append_3_out_aag₂(x1, x2)
if_append_3_in_1_aag₅(x1, x2, x3, x4, x5) = if_append_3_in_1_aag₂(x1, x5)
FL_3_IN_AGA₃(x1, x2, x3) = FL_3_IN_AGA₁(x2)
IF_FL_3_IN_1_AGA₅(x1, x2, x3, x4, x5) = IF_FL_3_IN_1_AGA₁(x5)

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

                        ↳ RuleRemovalProof

  ↳ PrologToPiTRSProof

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

IF_FL_3_IN_1_AGA₁(append_3_out_aag₂(E, Y)) -> FL_3_IN_AGA₁(Y)
FL_3_IN_AGA₁(R) -> IF_FL_3_IN_1_AGA₁(append_3_in_aag₁(R))

The TRS R consists of the following rules:

append_3_in_aag₁(X) -> append_3_out_aag₂([]_0, X)
append_3_in_aag₁(._2₂(X, Zs)) -> if_append_3_in_1_aag₂(X, append_3_in_aag₁(Zs))
if_append_3_in_1_aag₂(X, append_3_out_aag₂(Xs, Ys)) -> append_3_out_aag₂(._2₂(X, Xs), Ys)

The set Q consists of the following terms:

append_3_in_aag₁(x0)
if_append_3_in_1_aag₂(x0, x1)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {FL_3_IN_AGA₁, IF_FL_3_IN_1_AGA₁}.
By using a polynomial ordering, 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_3_in_aag₁(._2₂(X, Zs)) -> if_append_3_in_1_aag₂(X, append_3_in_aag₁(Zs))

Used ordering: POLO with Polynomial interpretation:

POL(FL_3_IN_AGA₁(x₁)) = 1 + 2·x₁
POL(append_3_in_aag₁(x₁)) = 1 + 2·x₁
POL(._2₂(x₁, x₂)) = 1 + x₁ + x₂
POL(append_3_out_aag₂(x₁, x₂)) = 1 + x₁ + 2·x₂
POL(if_append_3_in_1_aag₂(x₁, x₂)) = 1 + x₁ + x₂
POL(IF_FL_3_IN_1_AGA₁(x₁)) = x₁
POL([]_0) = 0

↳ 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:

IF_FL_3_IN_1_AGA₁(append_3_out_aag₂(E, Y)) -> FL_3_IN_AGA₁(Y)
FL_3_IN_AGA₁(R) -> IF_FL_3_IN_1_AGA₁(append_3_in_aag₁(R))

The TRS R consists of the following rules:

append_3_in_aag₁(X) -> append_3_out_aag₂([]_0, X)
if_append_3_in_1_aag₂(X, append_3_out_aag₂(Xs, Ys)) -> append_3_out_aag₂(._2₂(X, Xs), Ys)

The set Q consists of the following terms:

append_3_in_aag₁(x0)
if_append_3_in_1_aag₂(x0, x1)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {FL_3_IN_AGA₁, IF_FL_3_IN_1_AGA₁}.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, we can delete all non-usable rules from R.

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

  ↳ PrologToPiTRSProof

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

IF_FL_3_IN_1_AGA₁(append_3_out_aag₂(E, Y)) -> FL_3_IN_AGA₁(Y)
FL_3_IN_AGA₁(R) -> IF_FL_3_IN_1_AGA₁(append_3_in_aag₁(R))

The TRS R consists of the following rules:

append_3_in_aag₁(X) -> append_3_out_aag₂([]_0, X)

The set Q consists of the following terms:

append_3_in_aag₁(x0)
if_append_3_in_1_aag₂(x0, x1)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {FL_3_IN_AGA₁, IF_FL_3_IN_1_AGA₁}.
With regard to the inferred argument filtering the predicates were used in the following modes:
fl₃: (f,b,f)
append₃: (f,f,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

fl_3_in_aga₃([]_0, []_0, 0_0) -> fl_3_out_aga₃([]_0, []_0, 0_0)
fl_3_in_aga₃(._2₂(E, X), R, s_1₁(Z)) -> if_fl_3_in_1_aga₅(E, X, R, Z, append_3_in_aag₃(E, Y, R))
append_3_in_aag₃([]_0, X, X) -> append_3_out_aag₃([]_0, X, X)
append_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_in_aag₃(Xs, Ys, Zs))
if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_out_aag₃(Xs, Ys, Zs)) -> append_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_fl_3_in_1_aga₅(E, X, R, Z, append_3_out_aag₃(E, Y, R)) -> if_fl_3_in_2_aga₆(E, X, R, Z, Y, fl_3_in_aga₃(X, Y, Z))
if_fl_3_in_2_aga₆(E, X, R, Z, Y, fl_3_out_aga₃(X, Y, Z)) -> fl_3_out_aga₃(._2₂(E, X), R, s_1₁(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_3_in_aga₃([]_0, []_0, 0_0) -> fl_3_out_aga₃([]_0, []_0, 0_0)
fl_3_in_aga₃(._2₂(E, X), R, s_1₁(Z)) -> if_fl_3_in_1_aga₅(E, X, R, Z, append_3_in_aag₃(E, Y, R))
append_3_in_aag₃([]_0, X, X) -> append_3_out_aag₃([]_0, X, X)
append_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_in_aag₃(Xs, Ys, Zs))
if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_out_aag₃(Xs, Ys, Zs)) -> append_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_fl_3_in_1_aga₅(E, X, R, Z, append_3_out_aag₃(E, Y, R)) -> if_fl_3_in_2_aga₆(E, X, R, Z, Y, fl_3_in_aga₃(X, Y, Z))
if_fl_3_in_2_aga₆(E, X, R, Z, Y, fl_3_out_aga₃(X, Y, Z)) -> fl_3_out_aga₃(._2₂(E, X), R, s_1₁(Z))

FL_3_IN_AGA₃(._2₂(E, X), R, s_1₁(Z)) -> IF_FL_3_IN_1_AGA₅(E, X, R, Z, append_3_in_aag₃(E, Y, R))
FL_3_IN_AGA₃(._2₂(E, X), R, s_1₁(Z)) -> APPEND_3_IN_AAG₃(E, Y, R)
APPEND_3_IN_AAG₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> IF_APPEND_3_IN_1_AAG₅(X, Xs, Ys, Zs, append_3_in_aag₃(Xs, Ys, Zs))
APPEND_3_IN_AAG₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APPEND_3_IN_AAG₃(Xs, Ys, Zs)
IF_FL_3_IN_1_AGA₅(E, X, R, Z, append_3_out_aag₃(E, Y, R)) -> IF_FL_3_IN_2_AGA₆(E, X, R, Z, Y, fl_3_in_aga₃(X, Y, Z))
IF_FL_3_IN_1_AGA₅(E, X, R, Z, append_3_out_aag₃(E, Y, R)) -> FL_3_IN_AGA₃(X, Y, Z)

The TRS R consists of the following rules:

fl_3_in_aga₃([]_0, []_0, 0_0) -> fl_3_out_aga₃([]_0, []_0, 0_0)
fl_3_in_aga₃(._2₂(E, X), R, s_1₁(Z)) -> if_fl_3_in_1_aga₅(E, X, R, Z, append_3_in_aag₃(E, Y, R))
append_3_in_aag₃([]_0, X, X) -> append_3_out_aag₃([]_0, X, X)
append_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_in_aag₃(Xs, Ys, Zs))
if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_out_aag₃(Xs, Ys, Zs)) -> append_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_fl_3_in_1_aga₅(E, X, R, Z, append_3_out_aag₃(E, Y, R)) -> if_fl_3_in_2_aga₆(E, X, R, Z, Y, fl_3_in_aga₃(X, Y, Z))
if_fl_3_in_2_aga₆(E, X, R, Z, Y, fl_3_out_aga₃(X, Y, Z)) -> fl_3_out_aga₃(._2₂(E, X), R, s_1₁(Z))

The argument filtering Pi contains the following mapping:
fl_3_in_aga₃(x1, x2, x3) = fl_3_in_aga₁(x2)
[]_0 = []_0
0_0 = 0_0
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
fl_3_out_aga₃(x1, x2, x3) = fl_3_out_aga₃(x1, x2, x3)
if_fl_3_in_1_aga₅(x1, x2, x3, x4, x5) = if_fl_3_in_1_aga₂(x3, x5)
append_3_in_aag₃(x1, x2, x3) = append_3_in_aag₁(x3)
append_3_out_aag₃(x1, x2, x3) = append_3_out_aag₃(x1, x2, x3)
if_append_3_in_1_aag₅(x1, x2, x3, x4, x5) = if_append_3_in_1_aag₃(x1, x4, x5)
if_fl_3_in_2_aga₆(x1, x2, x3, x4, x5, x6) = if_fl_3_in_2_aga₃(x1, x3, x6)
IF_APPEND_3_IN_1_AAG₅(x1, x2, x3, x4, x5) = IF_APPEND_3_IN_1_AAG₃(x1, x4, x5)
FL_3_IN_AGA₃(x1, x2, x3) = FL_3_IN_AGA₁(x2)
APPEND_3_IN_AAG₃(x1, x2, x3) = APPEND_3_IN_AAG₁(x3)
IF_FL_3_IN_2_AGA₆(x1, x2, x3, x4, x5, x6) = IF_FL_3_IN_2_AGA₃(x1, x3, x6)
IF_FL_3_IN_1_AGA₅(x1, x2, x3, x4, x5) = IF_FL_3_IN_1_AGA₂(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_3_IN_AGA₃(._2₂(E, X), R, s_1₁(Z)) -> IF_FL_3_IN_1_AGA₅(E, X, R, Z, append_3_in_aag₃(E, Y, R))
FL_3_IN_AGA₃(._2₂(E, X), R, s_1₁(Z)) -> APPEND_3_IN_AAG₃(E, Y, R)
APPEND_3_IN_AAG₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> IF_APPEND_3_IN_1_AAG₅(X, Xs, Ys, Zs, append_3_in_aag₃(Xs, Ys, Zs))
APPEND_3_IN_AAG₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APPEND_3_IN_AAG₃(Xs, Ys, Zs)
IF_FL_3_IN_1_AGA₅(E, X, R, Z, append_3_out_aag₃(E, Y, R)) -> IF_FL_3_IN_2_AGA₆(E, X, R, Z, Y, fl_3_in_aga₃(X, Y, Z))
IF_FL_3_IN_1_AGA₅(E, X, R, Z, append_3_out_aag₃(E, Y, R)) -> FL_3_IN_AGA₃(X, Y, Z)

The TRS R consists of the following rules:

fl_3_in_aga₃([]_0, []_0, 0_0) -> fl_3_out_aga₃([]_0, []_0, 0_0)
fl_3_in_aga₃(._2₂(E, X), R, s_1₁(Z)) -> if_fl_3_in_1_aga₅(E, X, R, Z, append_3_in_aag₃(E, Y, R))
append_3_in_aag₃([]_0, X, X) -> append_3_out_aag₃([]_0, X, X)
append_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_in_aag₃(Xs, Ys, Zs))
if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_out_aag₃(Xs, Ys, Zs)) -> append_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_fl_3_in_1_aga₅(E, X, R, Z, append_3_out_aag₃(E, Y, R)) -> if_fl_3_in_2_aga₆(E, X, R, Z, Y, fl_3_in_aga₃(X, Y, Z))
if_fl_3_in_2_aga₆(E, X, R, Z, Y, fl_3_out_aga₃(X, Y, Z)) -> fl_3_out_aga₃(._2₂(E, X), R, s_1₁(Z))

The argument filtering Pi contains the following mapping:
fl_3_in_aga₃(x1, x2, x3) = fl_3_in_aga₁(x2)
[]_0 = []_0
0_0 = 0_0
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
fl_3_out_aga₃(x1, x2, x3) = fl_3_out_aga₃(x1, x2, x3)
if_fl_3_in_1_aga₅(x1, x2, x3, x4, x5) = if_fl_3_in_1_aga₂(x3, x5)
append_3_in_aag₃(x1, x2, x3) = append_3_in_aag₁(x3)
append_3_out_aag₃(x1, x2, x3) = append_3_out_aag₃(x1, x2, x3)
if_append_3_in_1_aag₅(x1, x2, x3, x4, x5) = if_append_3_in_1_aag₃(x1, x4, x5)
if_fl_3_in_2_aga₆(x1, x2, x3, x4, x5, x6) = if_fl_3_in_2_aga₃(x1, x3, x6)
IF_APPEND_3_IN_1_AAG₅(x1, x2, x3, x4, x5) = IF_APPEND_3_IN_1_AAG₃(x1, x4, x5)
FL_3_IN_AGA₃(x1, x2, x3) = FL_3_IN_AGA₁(x2)
APPEND_3_IN_AAG₃(x1, x2, x3) = APPEND_3_IN_AAG₁(x3)
IF_FL_3_IN_2_AGA₆(x1, x2, x3, x4, x5, x6) = IF_FL_3_IN_2_AGA₃(x1, x3, x6)
IF_FL_3_IN_1_AGA₅(x1, x2, x3, x4, x5) = IF_FL_3_IN_1_AGA₂(x3, x5)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph 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_3_IN_AAG₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APPEND_3_IN_AAG₃(Xs, Ys, Zs)

The TRS R consists of the following rules:

fl_3_in_aga₃([]_0, []_0, 0_0) -> fl_3_out_aga₃([]_0, []_0, 0_0)
fl_3_in_aga₃(._2₂(E, X), R, s_1₁(Z)) -> if_fl_3_in_1_aga₅(E, X, R, Z, append_3_in_aag₃(E, Y, R))
append_3_in_aag₃([]_0, X, X) -> append_3_out_aag₃([]_0, X, X)
append_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_in_aag₃(Xs, Ys, Zs))
if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_out_aag₃(Xs, Ys, Zs)) -> append_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_fl_3_in_1_aga₅(E, X, R, Z, append_3_out_aag₃(E, Y, R)) -> if_fl_3_in_2_aga₆(E, X, R, Z, Y, fl_3_in_aga₃(X, Y, Z))
if_fl_3_in_2_aga₆(E, X, R, Z, Y, fl_3_out_aga₃(X, Y, Z)) -> fl_3_out_aga₃(._2₂(E, X), R, s_1₁(Z))

The argument filtering Pi contains the following mapping:
fl_3_in_aga₃(x1, x2, x3) = fl_3_in_aga₁(x2)
[]_0 = []_0
0_0 = 0_0
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
fl_3_out_aga₃(x1, x2, x3) = fl_3_out_aga₃(x1, x2, x3)
if_fl_3_in_1_aga₅(x1, x2, x3, x4, x5) = if_fl_3_in_1_aga₂(x3, x5)
append_3_in_aag₃(x1, x2, x3) = append_3_in_aag₁(x3)
append_3_out_aag₃(x1, x2, x3) = append_3_out_aag₃(x1, x2, x3)
if_append_3_in_1_aag₅(x1, x2, x3, x4, x5) = if_append_3_in_1_aag₃(x1, x4, x5)
if_fl_3_in_2_aga₆(x1, x2, x3, x4, x5, x6) = if_fl_3_in_2_aga₃(x1, x3, x6)
APPEND_3_IN_AAG₃(x1, x2, x3) = APPEND_3_IN_AAG₁(x3)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting 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_3_IN_AAG₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APPEND_3_IN_AAG₃(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_3_IN_AAG₁(._2₂(X, Zs)) -> APPEND_3_IN_AAG₁(Zs)

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

APPEND_3_IN_AAG₁(._2₂(X, Zs)) -> APPEND_3_IN_AAG₁(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:

IF_FL_3_IN_1_AGA₅(E, X, R, Z, append_3_out_aag₃(E, Y, R)) -> FL_3_IN_AGA₃(X, Y, Z)
FL_3_IN_AGA₃(._2₂(E, X), R, s_1₁(Z)) -> IF_FL_3_IN_1_AGA₅(E, X, R, Z, append_3_in_aag₃(E, Y, R))

The TRS R consists of the following rules:

fl_3_in_aga₃([]_0, []_0, 0_0) -> fl_3_out_aga₃([]_0, []_0, 0_0)
fl_3_in_aga₃(._2₂(E, X), R, s_1₁(Z)) -> if_fl_3_in_1_aga₅(E, X, R, Z, append_3_in_aag₃(E, Y, R))
append_3_in_aag₃([]_0, X, X) -> append_3_out_aag₃([]_0, X, X)
append_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_in_aag₃(Xs, Ys, Zs))
if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_out_aag₃(Xs, Ys, Zs)) -> append_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_fl_3_in_1_aga₅(E, X, R, Z, append_3_out_aag₃(E, Y, R)) -> if_fl_3_in_2_aga₆(E, X, R, Z, Y, fl_3_in_aga₃(X, Y, Z))
if_fl_3_in_2_aga₆(E, X, R, Z, Y, fl_3_out_aga₃(X, Y, Z)) -> fl_3_out_aga₃(._2₂(E, X), R, s_1₁(Z))

The argument filtering Pi contains the following mapping:
fl_3_in_aga₃(x1, x2, x3) = fl_3_in_aga₁(x2)
[]_0 = []_0
0_0 = 0_0
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
fl_3_out_aga₃(x1, x2, x3) = fl_3_out_aga₃(x1, x2, x3)
if_fl_3_in_1_aga₅(x1, x2, x3, x4, x5) = if_fl_3_in_1_aga₂(x3, x5)
append_3_in_aag₃(x1, x2, x3) = append_3_in_aag₁(x3)
append_3_out_aag₃(x1, x2, x3) = append_3_out_aag₃(x1, x2, x3)
if_append_3_in_1_aag₅(x1, x2, x3, x4, x5) = if_append_3_in_1_aag₃(x1, x4, x5)
if_fl_3_in_2_aga₆(x1, x2, x3, x4, x5, x6) = if_fl_3_in_2_aga₃(x1, x3, x6)
FL_3_IN_AGA₃(x1, x2, x3) = FL_3_IN_AGA₁(x2)
IF_FL_3_IN_1_AGA₅(x1, x2, x3, x4, x5) = IF_FL_3_IN_1_AGA₂(x3, x5)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting 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:

IF_FL_3_IN_1_AGA₅(E, X, R, Z, append_3_out_aag₃(E, Y, R)) -> FL_3_IN_AGA₃(X, Y, Z)
FL_3_IN_AGA₃(._2₂(E, X), R, s_1₁(Z)) -> IF_FL_3_IN_1_AGA₅(E, X, R, Z, append_3_in_aag₃(E, Y, R))

The TRS R consists of the following rules:

append_3_in_aag₃([]_0, X, X) -> append_3_out_aag₃([]_0, X, X)
append_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_in_aag₃(Xs, Ys, Zs))
if_append_3_in_1_aag₅(X, Xs, Ys, Zs, append_3_out_aag₃(Xs, Ys, Zs)) -> append_3_out_aag₃(._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_aag₃(x1, x2, x3) = append_3_in_aag₁(x3)
append_3_out_aag₃(x1, x2, x3) = append_3_out_aag₃(x1, x2, x3)
if_append_3_in_1_aag₅(x1, x2, x3, x4, x5) = if_append_3_in_1_aag₃(x1, x4, x5)
FL_3_IN_AGA₃(x1, x2, x3) = FL_3_IN_AGA₁(x2)
IF_FL_3_IN_1_AGA₅(x1, x2, x3, x4, x5) = IF_FL_3_IN_1_AGA₂(x3, x5)

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

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

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

IF_FL_3_IN_1_AGA₂(R, append_3_out_aag₃(E, Y, R)) -> FL_3_IN_AGA₁(Y)
FL_3_IN_AGA₁(R) -> IF_FL_3_IN_1_AGA₂(R, append_3_in_aag₁(R))

The TRS R consists of the following rules:

append_3_in_aag₁(X) -> append_3_out_aag₃([]_0, X, X)
append_3_in_aag₁(._2₂(X, Zs)) -> if_append_3_in_1_aag₃(X, Zs, append_3_in_aag₁(Zs))
if_append_3_in_1_aag₃(X, Zs, append_3_out_aag₃(Xs, Ys, Zs)) -> append_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))

The set Q consists of the following terms:

append_3_in_aag₁(x0)
if_append_3_in_1_aag₃(x0, x1, x2)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {FL_3_IN_AGA₁, IF_FL_3_IN_1_AGA₂}.