Left Termination proof of /tmp/tmpS

Left Termination of the query pattern app3_b(b,b,b,f) w.r.t. the given Prolog program could successfully be proven:

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

app3_a₄(Xs, Ys, Zs, Us) :- app₃(Xs, Ys, Vs), app₃(Vs, Zs, Us).
app3_b₄(Xs, Ys, Zs, Us) :- app₃(Ys, Zs, Vs), app₃(Xs, Vs, Us).
app₃({}₀, Ys, Ys).
app₃(.₂(X, Xs), Ys, .₂(X, Zs)) :- app₃(Xs, Ys, Zs).

The clause

app3_a₄(Xs, Ys, Zs, Us) :- app₃(Xs, Ys, Vs), app₃(Vs, Zs, Us).

can be ignored, as it is not needed by any of the given querys.

Deleting this clauses results in the following prolog program:

app3_b₄(Xs, Ys, Zs, Us) :- app₃(Ys, Zs, Vs), app₃(Xs, Vs, Us).
app₃({}₀, Ys, Ys).
app₃(.₂(X, Xs), Ys, .₂(X, Zs)) :- app₃(Xs, Ys, Zs).

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

    ↳ PROLOG

      ↳ PrologToPiTRSProof

app3_b₄(Xs, Ys, Zs, Us) :- app₃(Ys, Zs, Vs), app₃(Xs, Vs, Us).
app₃({}₀, Ys, Ys).
app₃(.₂(X, Xs), Ys, .₂(X, Zs)) :- app₃(Xs, Ys, Zs).

With regard to the inferred argument filtering the predicates were used in the following modes:
app3_b₄: (b,b,b,f)
app₃: (b,b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

app3_b_4_in_ggga₄(Xs, Ys, Zs, Us) -> if_app3_b_4_in_1_ggga₅(Xs, Ys, Zs, Us, app_3_in_gga₃(Ys, Zs, Vs))
app_3_in_gga₃([]_0, Ys, Ys) -> app_3_out_gga₃([]_0, Ys, Ys)
app_3_in_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Xs, Ys, Zs)) -> app_3_out_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_app3_b_4_in_1_ggga₅(Xs, Ys, Zs, Us, app_3_out_gga₃(Ys, Zs, Vs)) -> if_app3_b_4_in_2_ggga₆(Xs, Ys, Zs, Us, Vs, app_3_in_gga₃(Xs, Vs, Us))
if_app3_b_4_in_2_ggga₆(Xs, Ys, Zs, Us, Vs, app_3_out_gga₃(Xs, Vs, Us)) -> app3_b_4_out_ggga₄(Xs, Ys, Zs, Us)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

    ↳ PROLOG

      ↳ PrologToPiTRSProof

        ↳ PiTRS

          ↳ DependencyPairsProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

app3_b_4_in_ggga₄(Xs, Ys, Zs, Us) -> if_app3_b_4_in_1_ggga₅(Xs, Ys, Zs, Us, app_3_in_gga₃(Ys, Zs, Vs))
app_3_in_gga₃([]_0, Ys, Ys) -> app_3_out_gga₃([]_0, Ys, Ys)
app_3_in_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Xs, Ys, Zs)) -> app_3_out_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_app3_b_4_in_1_ggga₅(Xs, Ys, Zs, Us, app_3_out_gga₃(Ys, Zs, Vs)) -> if_app3_b_4_in_2_ggga₆(Xs, Ys, Zs, Us, Vs, app_3_in_gga₃(Xs, Vs, Us))
if_app3_b_4_in_2_ggga₆(Xs, Ys, Zs, Us, Vs, app_3_out_gga₃(Xs, Vs, Us)) -> app3_b_4_out_ggga₄(Xs, Ys, Zs, Us)

APP3_B_4_IN_GGGA₄(Xs, Ys, Zs, Us) -> IF_APP3_B_4_IN_1_GGGA₅(Xs, Ys, Zs, Us, app_3_in_gga₃(Ys, Zs, Vs))
APP3_B_4_IN_GGGA₄(Xs, Ys, Zs, Us) -> APP_3_IN_GGA₃(Ys, Zs, Vs)
APP_3_IN_GGA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> IF_APP_3_IN_1_GGA₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
APP_3_IN_GGA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_GGA₃(Xs, Ys, Zs)
IF_APP3_B_4_IN_1_GGGA₅(Xs, Ys, Zs, Us, app_3_out_gga₃(Ys, Zs, Vs)) -> IF_APP3_B_4_IN_2_GGGA₆(Xs, Ys, Zs, Us, Vs, app_3_in_gga₃(Xs, Vs, Us))
IF_APP3_B_4_IN_1_GGGA₅(Xs, Ys, Zs, Us, app_3_out_gga₃(Ys, Zs, Vs)) -> APP_3_IN_GGA₃(Xs, Vs, Us)

The TRS R consists of the following rules:

app3_b_4_in_ggga₄(Xs, Ys, Zs, Us) -> if_app3_b_4_in_1_ggga₅(Xs, Ys, Zs, Us, app_3_in_gga₃(Ys, Zs, Vs))
app_3_in_gga₃([]_0, Ys, Ys) -> app_3_out_gga₃([]_0, Ys, Ys)
app_3_in_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Xs, Ys, Zs)) -> app_3_out_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_app3_b_4_in_1_ggga₅(Xs, Ys, Zs, Us, app_3_out_gga₃(Ys, Zs, Vs)) -> if_app3_b_4_in_2_ggga₆(Xs, Ys, Zs, Us, Vs, app_3_in_gga₃(Xs, Vs, Us))
if_app3_b_4_in_2_ggga₆(Xs, Ys, Zs, Us, Vs, app_3_out_gga₃(Xs, Vs, Us)) -> app3_b_4_out_ggga₄(Xs, Ys, Zs, Us)

The argument filtering Pi contains the following mapping:
app3_b_4_in_ggga₄(x1, x2, x3, x4) = app3_b_4_in_ggga₃(x1, x2, x3)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
if_app3_b_4_in_1_ggga₅(x1, x2, x3, x4, x5) = if_app3_b_4_in_1_ggga₂(x1, x5)
app_3_in_gga₃(x1, x2, x3) = app_3_in_gga₂(x1, x2)
app_3_out_gga₃(x1, x2, x3) = app_3_out_gga₁(x3)
if_app_3_in_1_gga₅(x1, x2, x3, x4, x5) = if_app_3_in_1_gga₂(x1, x5)
if_app3_b_4_in_2_ggga₆(x1, x2, x3, x4, x5, x6) = if_app3_b_4_in_2_ggga₁(x6)
app3_b_4_out_ggga₄(x1, x2, x3, x4) = app3_b_4_out_ggga₁(x4)
IF_APP3_B_4_IN_1_GGGA₅(x1, x2, x3, x4, x5) = IF_APP3_B_4_IN_1_GGGA₂(x1, x5)
IF_APP3_B_4_IN_2_GGGA₆(x1, x2, x3, x4, x5, x6) = IF_APP3_B_4_IN_2_GGGA₁(x6)
APP3_B_4_IN_GGGA₄(x1, x2, x3, x4) = APP3_B_4_IN_GGGA₃(x1, x2, x3)
IF_APP_3_IN_1_GGA₅(x1, x2, x3, x4, x5) = IF_APP_3_IN_1_GGA₂(x1, x5)
APP_3_IN_GGA₃(x1, x2, x3) = APP_3_IN_GGA₂(x1, x2)

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

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

    ↳ PROLOG

      ↳ PrologToPiTRSProof

        ↳ PiTRS

          ↳ DependencyPairsProof

            ↳ PiDP

              ↳ DependencyGraphProof

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

APP3_B_4_IN_GGGA₄(Xs, Ys, Zs, Us) -> IF_APP3_B_4_IN_1_GGGA₅(Xs, Ys, Zs, Us, app_3_in_gga₃(Ys, Zs, Vs))
APP3_B_4_IN_GGGA₄(Xs, Ys, Zs, Us) -> APP_3_IN_GGA₃(Ys, Zs, Vs)
APP_3_IN_GGA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> IF_APP_3_IN_1_GGA₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
APP_3_IN_GGA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_GGA₃(Xs, Ys, Zs)
IF_APP3_B_4_IN_1_GGGA₅(Xs, Ys, Zs, Us, app_3_out_gga₃(Ys, Zs, Vs)) -> IF_APP3_B_4_IN_2_GGGA₆(Xs, Ys, Zs, Us, Vs, app_3_in_gga₃(Xs, Vs, Us))
IF_APP3_B_4_IN_1_GGGA₅(Xs, Ys, Zs, Us, app_3_out_gga₃(Ys, Zs, Vs)) -> APP_3_IN_GGA₃(Xs, Vs, Us)

The TRS R consists of the following rules:

app3_b_4_in_ggga₄(Xs, Ys, Zs, Us) -> if_app3_b_4_in_1_ggga₅(Xs, Ys, Zs, Us, app_3_in_gga₃(Ys, Zs, Vs))
app_3_in_gga₃([]_0, Ys, Ys) -> app_3_out_gga₃([]_0, Ys, Ys)
app_3_in_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Xs, Ys, Zs)) -> app_3_out_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_app3_b_4_in_1_ggga₅(Xs, Ys, Zs, Us, app_3_out_gga₃(Ys, Zs, Vs)) -> if_app3_b_4_in_2_ggga₆(Xs, Ys, Zs, Us, Vs, app_3_in_gga₃(Xs, Vs, Us))
if_app3_b_4_in_2_ggga₆(Xs, Ys, Zs, Us, Vs, app_3_out_gga₃(Xs, Vs, Us)) -> app3_b_4_out_ggga₄(Xs, Ys, Zs, Us)

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

    ↳ PROLOG

      ↳ PrologToPiTRSProof

        ↳ PiTRS

          ↳ DependencyPairsProof

            ↳ PiDP

              ↳ DependencyGraphProof

                ↳ PiDP

                  ↳ UsableRulesProof

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

APP_3_IN_GGA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_GGA₃(Xs, Ys, Zs)

The TRS R consists of the following rules:

app3_b_4_in_ggga₄(Xs, Ys, Zs, Us) -> if_app3_b_4_in_1_ggga₅(Xs, Ys, Zs, Us, app_3_in_gga₃(Ys, Zs, Vs))
app_3_in_gga₃([]_0, Ys, Ys) -> app_3_out_gga₃([]_0, Ys, Ys)
app_3_in_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Xs, Ys, Zs)) -> app_3_out_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_app3_b_4_in_1_ggga₅(Xs, Ys, Zs, Us, app_3_out_gga₃(Ys, Zs, Vs)) -> if_app3_b_4_in_2_ggga₆(Xs, Ys, Zs, Us, Vs, app_3_in_gga₃(Xs, Vs, Us))
if_app3_b_4_in_2_ggga₆(Xs, Ys, Zs, Us, Vs, app_3_out_gga₃(Xs, Vs, Us)) -> app3_b_4_out_ggga₄(Xs, Ys, Zs, Us)

The argument filtering Pi contains the following mapping:
app3_b_4_in_ggga₄(x1, x2, x3, x4) = app3_b_4_in_ggga₃(x1, x2, x3)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
if_app3_b_4_in_1_ggga₅(x1, x2, x3, x4, x5) = if_app3_b_4_in_1_ggga₂(x1, x5)
app_3_in_gga₃(x1, x2, x3) = app_3_in_gga₂(x1, x2)
app_3_out_gga₃(x1, x2, x3) = app_3_out_gga₁(x3)
if_app_3_in_1_gga₅(x1, x2, x3, x4, x5) = if_app_3_in_1_gga₂(x1, x5)
if_app3_b_4_in_2_ggga₆(x1, x2, x3, x4, x5, x6) = if_app3_b_4_in_2_ggga₁(x6)
app3_b_4_out_ggga₄(x1, x2, x3, x4) = app3_b_4_out_ggga₁(x4)
APP_3_IN_GGA₃(x1, x2, x3) = APP_3_IN_GGA₂(x1, x2)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting we can delete all non-usable rules from R.

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

    ↳ PROLOG

      ↳ PrologToPiTRSProof

        ↳ PiTRS

          ↳ DependencyPairsProof

            ↳ PiDP

              ↳ DependencyGraphProof

                ↳ PiDP

                  ↳ UsableRulesProof

                    ↳ PiDP

                      ↳ PiDPToQDPProof

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

APP_3_IN_GGA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_GGA₃(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
APP_3_IN_GGA₃(x1, x2, x3) = APP_3_IN_GGA₂(x1, x2)

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

  ↳ UnrequestedClauseRemoverProof

    ↳ PROLOG

      ↳ PrologToPiTRSProof

        ↳ PiTRS

          ↳ DependencyPairsProof

            ↳ PiDP

              ↳ DependencyGraphProof

                ↳ PiDP

                  ↳ UsableRulesProof

                    ↳ PiDP

                      ↳ PiDPToQDPProof

                        ↳ QDP

                          ↳ QDPSizeChangeProof

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

APP_3_IN_GGA₂(._2₂(X, Xs), Ys) -> APP_3_IN_GGA₂(Xs, Ys)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {APP_3_IN_GGA₂}.
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:

APP_3_IN_GGA₂(._2₂(X, Xs), Ys) -> APP_3_IN_GGA₂(Xs, Ys)
The graph contains the following edges 1 > 1, 2 >= 2