Left Termination proof of /tmp/tmpGiw-SW/permutation-fb.pl

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

↳ PROLOG

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

perm₂({}₀, {}₀).
perm₂(Xs, .₂(X, Ys)) :- app₃(X1s, .₂(X, X2s), Xs), app₃(X1s, X2s, Zs), perm₂(Zs, Ys).
app₃({}₀, X, X).
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:
perm₂: (f,b)
app₃: (f,f,f) (b,f,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

perm_2_in_ag₂([]_0, []_0) -> perm_2_out_ag₂([]_0, []_0)
perm_2_in_ag₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_in_aaa₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aaa₃([]_0, X, X) -> app_3_out_aaa₃([]_0, X, X)
app_3_in_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_in_aaa₃(Xs, Ys, Zs))
if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_out_aaa₃(Xs, Ys, Zs)) -> app_3_out_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_in_gaa₃(X1s, X2s, Zs))
app_3_in_gaa₃([]_0, X, X) -> app_3_out_gaa₃([]_0, X, X)
app_3_in_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_in_gaa₃(Xs, Ys, Zs))
if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_out_gaa₃(Xs, Ys, Zs)) -> app_3_out_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_in_ag₂(Zs, Ys))
if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_out_ag₂(Zs, Ys)) -> perm_2_out_ag₂(Xs, ._2₂(X, Ys))

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:

perm_2_in_ag₂([]_0, []_0) -> perm_2_out_ag₂([]_0, []_0)
perm_2_in_ag₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_in_aaa₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aaa₃([]_0, X, X) -> app_3_out_aaa₃([]_0, X, X)
app_3_in_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_in_aaa₃(Xs, Ys, Zs))
if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_out_aaa₃(Xs, Ys, Zs)) -> app_3_out_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_in_gaa₃(X1s, X2s, Zs))
app_3_in_gaa₃([]_0, X, X) -> app_3_out_gaa₃([]_0, X, X)
app_3_in_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_in_gaa₃(Xs, Ys, Zs))
if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_out_gaa₃(Xs, Ys, Zs)) -> app_3_out_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_in_ag₂(Zs, Ys))
if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_out_ag₂(Zs, Ys)) -> perm_2_out_ag₂(Xs, ._2₂(X, Ys))

PERM_2_IN_AG₂(Xs, ._2₂(X, Ys)) -> IF_PERM_2_IN_1_AG₄(Xs, X, Ys, app_3_in_aaa₃(X1s, ._2₂(X, X2s), Xs))
PERM_2_IN_AG₂(Xs, ._2₂(X, Ys)) -> APP_3_IN_AAA₃(X1s, ._2₂(X, X2s), Xs)
APP_3_IN_AAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> IF_APP_3_IN_1_AAA₅(X, Xs, Ys, Zs, app_3_in_aaa₃(Xs, Ys, Zs))
APP_3_IN_AAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_AAA₃(Xs, Ys, Zs)
IF_PERM_2_IN_1_AG₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> IF_PERM_2_IN_2_AG₆(Xs, X, Ys, X1s, X2s, app_3_in_gaa₃(X1s, X2s, Zs))
IF_PERM_2_IN_1_AG₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> APP_3_IN_GAA₃(X1s, X2s, Zs)
APP_3_IN_GAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> IF_APP_3_IN_1_GAA₅(X, Xs, Ys, Zs, app_3_in_gaa₃(Xs, Ys, Zs))
APP_3_IN_GAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_GAA₃(Xs, Ys, Zs)
IF_PERM_2_IN_2_AG₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> IF_PERM_2_IN_3_AG₅(Xs, X, Ys, Zs, perm_2_in_ag₂(Zs, Ys))
IF_PERM_2_IN_2_AG₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> PERM_2_IN_AG₂(Zs, Ys)

The TRS R consists of the following rules:

perm_2_in_ag₂([]_0, []_0) -> perm_2_out_ag₂([]_0, []_0)
perm_2_in_ag₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_in_aaa₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aaa₃([]_0, X, X) -> app_3_out_aaa₃([]_0, X, X)
app_3_in_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_in_aaa₃(Xs, Ys, Zs))
if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_out_aaa₃(Xs, Ys, Zs)) -> app_3_out_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_in_gaa₃(X1s, X2s, Zs))
app_3_in_gaa₃([]_0, X, X) -> app_3_out_gaa₃([]_0, X, X)
app_3_in_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_in_gaa₃(Xs, Ys, Zs))
if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_out_gaa₃(Xs, Ys, Zs)) -> app_3_out_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_in_ag₂(Zs, Ys))
if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_out_ag₂(Zs, Ys)) -> perm_2_out_ag₂(Xs, ._2₂(X, Ys))

The argument filtering Pi contains the following mapping:
perm_2_in_ag₂(x1, x2) = perm_2_in_ag₁(x2)
[]_0 = []_0
._2₂(x1, x2) = ._2₁(x2)
perm_2_out_ag₂(x1, x2) = perm_2_out_ag
if_perm_2_in_1_ag₄(x1, x2, x3, x4) = if_perm_2_in_1_ag₂(x3, x4)
app_3_in_aaa₃(x1, x2, x3) = app_3_in_aaa
app_3_out_aaa₃(x1, x2, x3) = app_3_out_aaa₁(x1)
if_app_3_in_1_aaa₅(x1, x2, x3, x4, x5) = if_app_3_in_1_aaa₁(x5)
if_perm_2_in_2_ag₆(x1, x2, x3, x4, x5, x6) = if_perm_2_in_2_ag₂(x3, x6)
app_3_in_gaa₃(x1, x2, x3) = app_3_in_gaa₁(x1)
app_3_out_gaa₃(x1, x2, x3) = app_3_out_gaa
if_app_3_in_1_gaa₅(x1, x2, x3, x4, x5) = if_app_3_in_1_gaa₁(x5)
if_perm_2_in_3_ag₅(x1, x2, x3, x4, x5) = if_perm_2_in_3_ag₁(x5)
IF_APP_3_IN_1_GAA₅(x1, x2, x3, x4, x5) = IF_APP_3_IN_1_GAA₁(x5)
IF_PERM_2_IN_3_AG₅(x1, x2, x3, x4, x5) = IF_PERM_2_IN_3_AG₁(x5)
PERM_2_IN_AG₂(x1, x2) = PERM_2_IN_AG₁(x2)
IF_PERM_2_IN_2_AG₆(x1, x2, x3, x4, x5, x6) = IF_PERM_2_IN_2_AG₂(x3, x6)
IF_PERM_2_IN_1_AG₄(x1, x2, x3, x4) = IF_PERM_2_IN_1_AG₂(x3, x4)
APP_3_IN_GAA₃(x1, x2, x3) = APP_3_IN_GAA₁(x1)
APP_3_IN_AAA₃(x1, x2, x3) = APP_3_IN_AAA
IF_APP_3_IN_1_AAA₅(x1, x2, x3, x4, x5) = IF_APP_3_IN_1_AAA₁(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:

PERM_2_IN_AG₂(Xs, ._2₂(X, Ys)) -> IF_PERM_2_IN_1_AG₄(Xs, X, Ys, app_3_in_aaa₃(X1s, ._2₂(X, X2s), Xs))
PERM_2_IN_AG₂(Xs, ._2₂(X, Ys)) -> APP_3_IN_AAA₃(X1s, ._2₂(X, X2s), Xs)
APP_3_IN_AAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> IF_APP_3_IN_1_AAA₅(X, Xs, Ys, Zs, app_3_in_aaa₃(Xs, Ys, Zs))
APP_3_IN_AAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_AAA₃(Xs, Ys, Zs)
IF_PERM_2_IN_1_AG₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> IF_PERM_2_IN_2_AG₆(Xs, X, Ys, X1s, X2s, app_3_in_gaa₃(X1s, X2s, Zs))
IF_PERM_2_IN_1_AG₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> APP_3_IN_GAA₃(X1s, X2s, Zs)
APP_3_IN_GAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> IF_APP_3_IN_1_GAA₅(X, Xs, Ys, Zs, app_3_in_gaa₃(Xs, Ys, Zs))
APP_3_IN_GAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_GAA₃(Xs, Ys, Zs)
IF_PERM_2_IN_2_AG₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> IF_PERM_2_IN_3_AG₅(Xs, X, Ys, Zs, perm_2_in_ag₂(Zs, Ys))
IF_PERM_2_IN_2_AG₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> PERM_2_IN_AG₂(Zs, Ys)

The TRS R consists of the following rules:

perm_2_in_ag₂([]_0, []_0) -> perm_2_out_ag₂([]_0, []_0)
perm_2_in_ag₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_in_aaa₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aaa₃([]_0, X, X) -> app_3_out_aaa₃([]_0, X, X)
app_3_in_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_in_aaa₃(Xs, Ys, Zs))
if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_out_aaa₃(Xs, Ys, Zs)) -> app_3_out_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_in_gaa₃(X1s, X2s, Zs))
app_3_in_gaa₃([]_0, X, X) -> app_3_out_gaa₃([]_0, X, X)
app_3_in_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_in_gaa₃(Xs, Ys, Zs))
if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_out_gaa₃(Xs, Ys, Zs)) -> app_3_out_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_in_ag₂(Zs, Ys))
if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_out_ag₂(Zs, Ys)) -> perm_2_out_ag₂(Xs, ._2₂(X, Ys))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

APP_3_IN_GAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_GAA₃(Xs, Ys, Zs)

The TRS R consists of the following rules:

perm_2_in_ag₂([]_0, []_0) -> perm_2_out_ag₂([]_0, []_0)
perm_2_in_ag₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_in_aaa₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aaa₃([]_0, X, X) -> app_3_out_aaa₃([]_0, X, X)
app_3_in_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_in_aaa₃(Xs, Ys, Zs))
if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_out_aaa₃(Xs, Ys, Zs)) -> app_3_out_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_in_gaa₃(X1s, X2s, Zs))
app_3_in_gaa₃([]_0, X, X) -> app_3_out_gaa₃([]_0, X, X)
app_3_in_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_in_gaa₃(Xs, Ys, Zs))
if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_out_gaa₃(Xs, Ys, Zs)) -> app_3_out_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_in_ag₂(Zs, Ys))
if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_out_ag₂(Zs, Ys)) -> perm_2_out_ag₂(Xs, ._2₂(X, Ys))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

APP_3_IN_GAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_GAA₃(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₁(x2)
APP_3_IN_GAA₃(x1, x2, x3) = APP_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

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

APP_3_IN_GAA₁(._2₁(Xs)) -> APP_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 {APP_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:

APP_3_IN_GAA₁(._2₁(Xs)) -> APP_3_IN_GAA₁(Xs)
The graph contains the following edges 1 > 1

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

APP_3_IN_AAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_AAA₃(Xs, Ys, Zs)

The TRS R consists of the following rules:

perm_2_in_ag₂([]_0, []_0) -> perm_2_out_ag₂([]_0, []_0)
perm_2_in_ag₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_in_aaa₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aaa₃([]_0, X, X) -> app_3_out_aaa₃([]_0, X, X)
app_3_in_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_in_aaa₃(Xs, Ys, Zs))
if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_out_aaa₃(Xs, Ys, Zs)) -> app_3_out_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_in_gaa₃(X1s, X2s, Zs))
app_3_in_gaa₃([]_0, X, X) -> app_3_out_gaa₃([]_0, X, X)
app_3_in_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_in_gaa₃(Xs, Ys, Zs))
if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_out_gaa₃(Xs, Ys, Zs)) -> app_3_out_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_in_ag₂(Zs, Ys))
if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_out_ag₂(Zs, Ys)) -> perm_2_out_ag₂(Xs, ._2₂(X, Ys))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

APP_3_IN_AAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_AAA₃(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₁(x2)
APP_3_IN_AAA₃(x1, x2, x3) = APP_3_IN_AAA

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

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

APP_3_IN_AAA -> APP_3_IN_AAA

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_AAA}.

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

  ↳ PrologToPiTRSProof

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

PERM_2_IN_AG₂(Xs, ._2₂(X, Ys)) -> IF_PERM_2_IN_1_AG₄(Xs, X, Ys, app_3_in_aaa₃(X1s, ._2₂(X, X2s), Xs))
IF_PERM_2_IN_1_AG₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> IF_PERM_2_IN_2_AG₆(Xs, X, Ys, X1s, X2s, app_3_in_gaa₃(X1s, X2s, Zs))
IF_PERM_2_IN_2_AG₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> PERM_2_IN_AG₂(Zs, Ys)

The TRS R consists of the following rules:

perm_2_in_ag₂([]_0, []_0) -> perm_2_out_ag₂([]_0, []_0)
perm_2_in_ag₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_in_aaa₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aaa₃([]_0, X, X) -> app_3_out_aaa₃([]_0, X, X)
app_3_in_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_in_aaa₃(Xs, Ys, Zs))
if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_out_aaa₃(Xs, Ys, Zs)) -> app_3_out_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_in_gaa₃(X1s, X2s, Zs))
app_3_in_gaa₃([]_0, X, X) -> app_3_out_gaa₃([]_0, X, X)
app_3_in_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_in_gaa₃(Xs, Ys, Zs))
if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_out_gaa₃(Xs, Ys, Zs)) -> app_3_out_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_in_ag₂(Zs, Ys))
if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_out_ag₂(Zs, Ys)) -> perm_2_out_ag₂(Xs, ._2₂(X, Ys))

The argument filtering Pi contains the following mapping:
perm_2_in_ag₂(x1, x2) = perm_2_in_ag₁(x2)
[]_0 = []_0
._2₂(x1, x2) = ._2₁(x2)
perm_2_out_ag₂(x1, x2) = perm_2_out_ag
if_perm_2_in_1_ag₄(x1, x2, x3, x4) = if_perm_2_in_1_ag₂(x3, x4)
app_3_in_aaa₃(x1, x2, x3) = app_3_in_aaa
app_3_out_aaa₃(x1, x2, x3) = app_3_out_aaa₁(x1)
if_app_3_in_1_aaa₅(x1, x2, x3, x4, x5) = if_app_3_in_1_aaa₁(x5)
if_perm_2_in_2_ag₆(x1, x2, x3, x4, x5, x6) = if_perm_2_in_2_ag₂(x3, x6)
app_3_in_gaa₃(x1, x2, x3) = app_3_in_gaa₁(x1)
app_3_out_gaa₃(x1, x2, x3) = app_3_out_gaa
if_app_3_in_1_gaa₅(x1, x2, x3, x4, x5) = if_app_3_in_1_gaa₁(x5)
if_perm_2_in_3_ag₅(x1, x2, x3, x4, x5) = if_perm_2_in_3_ag₁(x5)
PERM_2_IN_AG₂(x1, x2) = PERM_2_IN_AG₁(x2)
IF_PERM_2_IN_2_AG₆(x1, x2, x3, x4, x5, x6) = IF_PERM_2_IN_2_AG₂(x3, x6)
IF_PERM_2_IN_1_AG₄(x1, x2, x3, x4) = IF_PERM_2_IN_1_AG₂(x3, x4)

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

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

  ↳ PrologToPiTRSProof

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

PERM_2_IN_AG₂(Xs, ._2₂(X, Ys)) -> IF_PERM_2_IN_1_AG₄(Xs, X, Ys, app_3_in_aaa₃(X1s, ._2₂(X, X2s), Xs))
IF_PERM_2_IN_1_AG₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> IF_PERM_2_IN_2_AG₆(Xs, X, Ys, X1s, X2s, app_3_in_gaa₃(X1s, X2s, Zs))
IF_PERM_2_IN_2_AG₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> PERM_2_IN_AG₂(Zs, Ys)

The TRS R consists of the following rules:

app_3_in_aaa₃([]_0, X, X) -> app_3_out_aaa₃([]_0, X, X)
app_3_in_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_in_aaa₃(Xs, Ys, Zs))
app_3_in_gaa₃([]_0, X, X) -> app_3_out_gaa₃([]_0, X, X)
app_3_in_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_in_gaa₃(Xs, Ys, Zs))
if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_out_aaa₃(Xs, Ys, Zs)) -> app_3_out_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_out_gaa₃(Xs, Ys, Zs)) -> app_3_out_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))

The argument filtering Pi contains the following mapping:
[]_0 = []_0
._2₂(x1, x2) = ._2₁(x2)
app_3_in_aaa₃(x1, x2, x3) = app_3_in_aaa
app_3_out_aaa₃(x1, x2, x3) = app_3_out_aaa₁(x1)
if_app_3_in_1_aaa₅(x1, x2, x3, x4, x5) = if_app_3_in_1_aaa₁(x5)
app_3_in_gaa₃(x1, x2, x3) = app_3_in_gaa₁(x1)
app_3_out_gaa₃(x1, x2, x3) = app_3_out_gaa
if_app_3_in_1_gaa₅(x1, x2, x3, x4, x5) = if_app_3_in_1_gaa₁(x5)
PERM_2_IN_AG₂(x1, x2) = PERM_2_IN_AG₁(x2)
IF_PERM_2_IN_2_AG₆(x1, x2, x3, x4, x5, x6) = IF_PERM_2_IN_2_AG₂(x3, x6)
IF_PERM_2_IN_1_AG₄(x1, x2, x3, x4) = IF_PERM_2_IN_1_AG₂(x3, x4)

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

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

  ↳ PrologToPiTRSProof

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

PERM_2_IN_AG₁(._2₁(Ys)) -> IF_PERM_2_IN_1_AG₂(Ys, app_3_in_aaa)
IF_PERM_2_IN_1_AG₂(Ys, app_3_out_aaa₁(X1s)) -> IF_PERM_2_IN_2_AG₂(Ys, app_3_in_gaa₁(X1s))
IF_PERM_2_IN_2_AG₂(Ys, app_3_out_gaa) -> PERM_2_IN_AG₁(Ys)

The TRS R consists of the following rules:

app_3_in_aaa -> app_3_out_aaa₁([]_0)
app_3_in_aaa -> if_app_3_in_1_aaa₁(app_3_in_aaa)
app_3_in_gaa₁([]_0) -> app_3_out_gaa
app_3_in_gaa₁(._2₁(Xs)) -> if_app_3_in_1_gaa₁(app_3_in_gaa₁(Xs))
if_app_3_in_1_aaa₁(app_3_out_aaa₁(Xs)) -> app_3_out_aaa₁(._2₁(Xs))
if_app_3_in_1_gaa₁(app_3_out_gaa) -> app_3_out_gaa

The set Q consists of the following terms:

app_3_in_aaa
app_3_in_gaa₁(x0)
if_app_3_in_1_aaa₁(x0)
if_app_3_in_1_gaa₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_PERM_2_IN_1_AG₂, PERM_2_IN_AG₁, IF_PERM_2_IN_2_AG₂}.
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_PERM_2_IN_1_AG₂(Ys, app_3_out_aaa₁(X1s)) -> IF_PERM_2_IN_2_AG₂(Ys, app_3_in_gaa₁(X1s))
The graph contains the following edges 1 >= 1
IF_PERM_2_IN_2_AG₂(Ys, app_3_out_gaa) -> PERM_2_IN_AG₁(Ys)
The graph contains the following edges 1 >= 1
PERM_2_IN_AG₁(._2₁(Ys)) -> IF_PERM_2_IN_1_AG₂(Ys, app_3_in_aaa)
The graph contains the following edges 1 > 1

perm_2_in_ag₂([]_0, []_0) -> perm_2_out_ag₂([]_0, []_0)
perm_2_in_ag₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_in_aaa₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aaa₃([]_0, X, X) -> app_3_out_aaa₃([]_0, X, X)
app_3_in_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_in_aaa₃(Xs, Ys, Zs))
if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_out_aaa₃(Xs, Ys, Zs)) -> app_3_out_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_in_gaa₃(X1s, X2s, Zs))
app_3_in_gaa₃([]_0, X, X) -> app_3_out_gaa₃([]_0, X, X)
app_3_in_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_in_gaa₃(Xs, Ys, Zs))
if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_out_gaa₃(Xs, Ys, Zs)) -> app_3_out_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_in_ag₂(Zs, Ys))
if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_out_ag₂(Zs, Ys)) -> perm_2_out_ag₂(Xs, ._2₂(X, Ys))

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:

perm_2_in_ag₂([]_0, []_0) -> perm_2_out_ag₂([]_0, []_0)
perm_2_in_ag₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_in_aaa₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aaa₃([]_0, X, X) -> app_3_out_aaa₃([]_0, X, X)
app_3_in_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_in_aaa₃(Xs, Ys, Zs))
if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_out_aaa₃(Xs, Ys, Zs)) -> app_3_out_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_in_gaa₃(X1s, X2s, Zs))
app_3_in_gaa₃([]_0, X, X) -> app_3_out_gaa₃([]_0, X, X)
app_3_in_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_in_gaa₃(Xs, Ys, Zs))
if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_out_gaa₃(Xs, Ys, Zs)) -> app_3_out_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_in_ag₂(Zs, Ys))
if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_out_ag₂(Zs, Ys)) -> perm_2_out_ag₂(Xs, ._2₂(X, Ys))

PERM_2_IN_AG₂(Xs, ._2₂(X, Ys)) -> IF_PERM_2_IN_1_AG₄(Xs, X, Ys, app_3_in_aaa₃(X1s, ._2₂(X, X2s), Xs))
PERM_2_IN_AG₂(Xs, ._2₂(X, Ys)) -> APP_3_IN_AAA₃(X1s, ._2₂(X, X2s), Xs)
APP_3_IN_AAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> IF_APP_3_IN_1_AAA₅(X, Xs, Ys, Zs, app_3_in_aaa₃(Xs, Ys, Zs))
APP_3_IN_AAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_AAA₃(Xs, Ys, Zs)
IF_PERM_2_IN_1_AG₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> IF_PERM_2_IN_2_AG₆(Xs, X, Ys, X1s, X2s, app_3_in_gaa₃(X1s, X2s, Zs))
IF_PERM_2_IN_1_AG₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> APP_3_IN_GAA₃(X1s, X2s, Zs)
APP_3_IN_GAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> IF_APP_3_IN_1_GAA₅(X, Xs, Ys, Zs, app_3_in_gaa₃(Xs, Ys, Zs))
APP_3_IN_GAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_GAA₃(Xs, Ys, Zs)
IF_PERM_2_IN_2_AG₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> IF_PERM_2_IN_3_AG₅(Xs, X, Ys, Zs, perm_2_in_ag₂(Zs, Ys))
IF_PERM_2_IN_2_AG₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> PERM_2_IN_AG₂(Zs, Ys)

The TRS R consists of the following rules:

perm_2_in_ag₂([]_0, []_0) -> perm_2_out_ag₂([]_0, []_0)
perm_2_in_ag₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_in_aaa₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aaa₃([]_0, X, X) -> app_3_out_aaa₃([]_0, X, X)
app_3_in_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_in_aaa₃(Xs, Ys, Zs))
if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_out_aaa₃(Xs, Ys, Zs)) -> app_3_out_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_in_gaa₃(X1s, X2s, Zs))
app_3_in_gaa₃([]_0, X, X) -> app_3_out_gaa₃([]_0, X, X)
app_3_in_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_in_gaa₃(Xs, Ys, Zs))
if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_out_gaa₃(Xs, Ys, Zs)) -> app_3_out_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_in_ag₂(Zs, Ys))
if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_out_ag₂(Zs, Ys)) -> perm_2_out_ag₂(Xs, ._2₂(X, Ys))

The argument filtering Pi contains the following mapping:
perm_2_in_ag₂(x1, x2) = perm_2_in_ag₁(x2)
[]_0 = []_0
._2₂(x1, x2) = ._2₁(x2)
perm_2_out_ag₂(x1, x2) = perm_2_out_ag₁(x2)
if_perm_2_in_1_ag₄(x1, x2, x3, x4) = if_perm_2_in_1_ag₂(x3, x4)
app_3_in_aaa₃(x1, x2, x3) = app_3_in_aaa
app_3_out_aaa₃(x1, x2, x3) = app_3_out_aaa₁(x1)
if_app_3_in_1_aaa₅(x1, x2, x3, x4, x5) = if_app_3_in_1_aaa₁(x5)
if_perm_2_in_2_ag₆(x1, x2, x3, x4, x5, x6) = if_perm_2_in_2_ag₂(x3, x6)
app_3_in_gaa₃(x1, x2, x3) = app_3_in_gaa₁(x1)
app_3_out_gaa₃(x1, x2, x3) = app_3_out_gaa₁(x1)
if_app_3_in_1_gaa₅(x1, x2, x3, x4, x5) = if_app_3_in_1_gaa₂(x2, x5)
if_perm_2_in_3_ag₅(x1, x2, x3, x4, x5) = if_perm_2_in_3_ag₂(x3, x5)
IF_APP_3_IN_1_GAA₅(x1, x2, x3, x4, x5) = IF_APP_3_IN_1_GAA₂(x2, x5)
IF_PERM_2_IN_3_AG₅(x1, x2, x3, x4, x5) = IF_PERM_2_IN_3_AG₂(x3, x5)
PERM_2_IN_AG₂(x1, x2) = PERM_2_IN_AG₁(x2)
IF_PERM_2_IN_2_AG₆(x1, x2, x3, x4, x5, x6) = IF_PERM_2_IN_2_AG₂(x3, x6)
IF_PERM_2_IN_1_AG₄(x1, x2, x3, x4) = IF_PERM_2_IN_1_AG₂(x3, x4)
APP_3_IN_GAA₃(x1, x2, x3) = APP_3_IN_GAA₁(x1)
APP_3_IN_AAA₃(x1, x2, x3) = APP_3_IN_AAA
IF_APP_3_IN_1_AAA₅(x1, x2, x3, x4, x5) = IF_APP_3_IN_1_AAA₁(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:

PERM_2_IN_AG₂(Xs, ._2₂(X, Ys)) -> IF_PERM_2_IN_1_AG₄(Xs, X, Ys, app_3_in_aaa₃(X1s, ._2₂(X, X2s), Xs))
PERM_2_IN_AG₂(Xs, ._2₂(X, Ys)) -> APP_3_IN_AAA₃(X1s, ._2₂(X, X2s), Xs)
APP_3_IN_AAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> IF_APP_3_IN_1_AAA₅(X, Xs, Ys, Zs, app_3_in_aaa₃(Xs, Ys, Zs))
APP_3_IN_AAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_AAA₃(Xs, Ys, Zs)
IF_PERM_2_IN_1_AG₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> IF_PERM_2_IN_2_AG₆(Xs, X, Ys, X1s, X2s, app_3_in_gaa₃(X1s, X2s, Zs))
IF_PERM_2_IN_1_AG₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> APP_3_IN_GAA₃(X1s, X2s, Zs)
APP_3_IN_GAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> IF_APP_3_IN_1_GAA₅(X, Xs, Ys, Zs, app_3_in_gaa₃(Xs, Ys, Zs))
APP_3_IN_GAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_GAA₃(Xs, Ys, Zs)
IF_PERM_2_IN_2_AG₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> IF_PERM_2_IN_3_AG₅(Xs, X, Ys, Zs, perm_2_in_ag₂(Zs, Ys))
IF_PERM_2_IN_2_AG₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> PERM_2_IN_AG₂(Zs, Ys)

The TRS R consists of the following rules:

perm_2_in_ag₂([]_0, []_0) -> perm_2_out_ag₂([]_0, []_0)
perm_2_in_ag₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_in_aaa₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aaa₃([]_0, X, X) -> app_3_out_aaa₃([]_0, X, X)
app_3_in_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_in_aaa₃(Xs, Ys, Zs))
if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_out_aaa₃(Xs, Ys, Zs)) -> app_3_out_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_in_gaa₃(X1s, X2s, Zs))
app_3_in_gaa₃([]_0, X, X) -> app_3_out_gaa₃([]_0, X, X)
app_3_in_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_in_gaa₃(Xs, Ys, Zs))
if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_out_gaa₃(Xs, Ys, Zs)) -> app_3_out_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_in_ag₂(Zs, Ys))
if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_out_ag₂(Zs, Ys)) -> perm_2_out_ag₂(Xs, ._2₂(X, Ys))

The argument filtering Pi contains the following mapping:
perm_2_in_ag₂(x1, x2) = perm_2_in_ag₁(x2)
[]_0 = []_0
._2₂(x1, x2) = ._2₁(x2)
perm_2_out_ag₂(x1, x2) = perm_2_out_ag₁(x2)
if_perm_2_in_1_ag₄(x1, x2, x3, x4) = if_perm_2_in_1_ag₂(x3, x4)
app_3_in_aaa₃(x1, x2, x3) = app_3_in_aaa
app_3_out_aaa₃(x1, x2, x3) = app_3_out_aaa₁(x1)
if_app_3_in_1_aaa₅(x1, x2, x3, x4, x5) = if_app_3_in_1_aaa₁(x5)
if_perm_2_in_2_ag₆(x1, x2, x3, x4, x5, x6) = if_perm_2_in_2_ag₂(x3, x6)
app_3_in_gaa₃(x1, x2, x3) = app_3_in_gaa₁(x1)
app_3_out_gaa₃(x1, x2, x3) = app_3_out_gaa₁(x1)
if_app_3_in_1_gaa₅(x1, x2, x3, x4, x5) = if_app_3_in_1_gaa₂(x2, x5)
if_perm_2_in_3_ag₅(x1, x2, x3, x4, x5) = if_perm_2_in_3_ag₂(x3, x5)
IF_APP_3_IN_1_GAA₅(x1, x2, x3, x4, x5) = IF_APP_3_IN_1_GAA₂(x2, x5)
IF_PERM_2_IN_3_AG₅(x1, x2, x3, x4, x5) = IF_PERM_2_IN_3_AG₂(x3, x5)
PERM_2_IN_AG₂(x1, x2) = PERM_2_IN_AG₁(x2)
IF_PERM_2_IN_2_AG₆(x1, x2, x3, x4, x5, x6) = IF_PERM_2_IN_2_AG₂(x3, x6)
IF_PERM_2_IN_1_AG₄(x1, x2, x3, x4) = IF_PERM_2_IN_1_AG₂(x3, x4)
APP_3_IN_GAA₃(x1, x2, x3) = APP_3_IN_GAA₁(x1)
APP_3_IN_AAA₃(x1, x2, x3) = APP_3_IN_AAA
IF_APP_3_IN_1_AAA₅(x1, x2, x3, x4, x5) = IF_APP_3_IN_1_AAA₁(x5)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 3 SCCs with 5 less nodes.

↳ PROLOG

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

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

APP_3_IN_GAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_GAA₃(Xs, Ys, Zs)

The TRS R consists of the following rules:

perm_2_in_ag₂([]_0, []_0) -> perm_2_out_ag₂([]_0, []_0)
perm_2_in_ag₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_in_aaa₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aaa₃([]_0, X, X) -> app_3_out_aaa₃([]_0, X, X)
app_3_in_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_in_aaa₃(Xs, Ys, Zs))
if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_out_aaa₃(Xs, Ys, Zs)) -> app_3_out_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_in_gaa₃(X1s, X2s, Zs))
app_3_in_gaa₃([]_0, X, X) -> app_3_out_gaa₃([]_0, X, X)
app_3_in_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_in_gaa₃(Xs, Ys, Zs))
if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_out_gaa₃(Xs, Ys, Zs)) -> app_3_out_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_in_ag₂(Zs, Ys))
if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_out_ag₂(Zs, Ys)) -> perm_2_out_ag₂(Xs, ._2₂(X, Ys))

↳ PROLOG

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

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

APP_3_IN_GAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_GAA₃(Xs, Ys, Zs)

↳ PROLOG

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

              ↳ PiDP

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

APP_3_IN_GAA₁(._2₁(Xs)) -> APP_3_IN_GAA₁(Xs)

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

APP_3_IN_GAA₁(._2₁(Xs)) -> APP_3_IN_GAA₁(Xs)
The graph contains the following edges 1 > 1

↳ PROLOG

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

APP_3_IN_AAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_AAA₃(Xs, Ys, Zs)

The TRS R consists of the following rules:

perm_2_in_ag₂([]_0, []_0) -> perm_2_out_ag₂([]_0, []_0)
perm_2_in_ag₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_in_aaa₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aaa₃([]_0, X, X) -> app_3_out_aaa₃([]_0, X, X)
app_3_in_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_in_aaa₃(Xs, Ys, Zs))
if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_out_aaa₃(Xs, Ys, Zs)) -> app_3_out_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_in_gaa₃(X1s, X2s, Zs))
app_3_in_gaa₃([]_0, X, X) -> app_3_out_gaa₃([]_0, X, X)
app_3_in_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_in_gaa₃(Xs, Ys, Zs))
if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_out_gaa₃(Xs, Ys, Zs)) -> app_3_out_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_in_ag₂(Zs, Ys))
if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_out_ag₂(Zs, Ys)) -> perm_2_out_ag₂(Xs, ._2₂(X, Ys))

↳ PROLOG

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

APP_3_IN_AAA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_AAA₃(Xs, Ys, Zs)

↳ PROLOG

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

              ↳ PiDP

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

APP_3_IN_AAA -> APP_3_IN_AAA

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_AAA}.

↳ PROLOG

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

PERM_2_IN_AG₂(Xs, ._2₂(X, Ys)) -> IF_PERM_2_IN_1_AG₄(Xs, X, Ys, app_3_in_aaa₃(X1s, ._2₂(X, X2s), Xs))
IF_PERM_2_IN_1_AG₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> IF_PERM_2_IN_2_AG₆(Xs, X, Ys, X1s, X2s, app_3_in_gaa₃(X1s, X2s, Zs))
IF_PERM_2_IN_2_AG₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> PERM_2_IN_AG₂(Zs, Ys)

The TRS R consists of the following rules:

perm_2_in_ag₂([]_0, []_0) -> perm_2_out_ag₂([]_0, []_0)
perm_2_in_ag₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_in_aaa₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aaa₃([]_0, X, X) -> app_3_out_aaa₃([]_0, X, X)
app_3_in_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_in_aaa₃(Xs, Ys, Zs))
if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_out_aaa₃(Xs, Ys, Zs)) -> app_3_out_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ag₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_in_gaa₃(X1s, X2s, Zs))
app_3_in_gaa₃([]_0, X, X) -> app_3_out_gaa₃([]_0, X, X)
app_3_in_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_in_gaa₃(Xs, Ys, Zs))
if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_out_gaa₃(Xs, Ys, Zs)) -> app_3_out_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_2_ag₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_in_ag₂(Zs, Ys))
if_perm_2_in_3_ag₅(Xs, X, Ys, Zs, perm_2_out_ag₂(Zs, Ys)) -> perm_2_out_ag₂(Xs, ._2₂(X, Ys))

The argument filtering Pi contains the following mapping:
perm_2_in_ag₂(x1, x2) = perm_2_in_ag₁(x2)
[]_0 = []_0
._2₂(x1, x2) = ._2₁(x2)
perm_2_out_ag₂(x1, x2) = perm_2_out_ag₁(x2)
if_perm_2_in_1_ag₄(x1, x2, x3, x4) = if_perm_2_in_1_ag₂(x3, x4)
app_3_in_aaa₃(x1, x2, x3) = app_3_in_aaa
app_3_out_aaa₃(x1, x2, x3) = app_3_out_aaa₁(x1)
if_app_3_in_1_aaa₅(x1, x2, x3, x4, x5) = if_app_3_in_1_aaa₁(x5)
if_perm_2_in_2_ag₆(x1, x2, x3, x4, x5, x6) = if_perm_2_in_2_ag₂(x3, x6)
app_3_in_gaa₃(x1, x2, x3) = app_3_in_gaa₁(x1)
app_3_out_gaa₃(x1, x2, x3) = app_3_out_gaa₁(x1)
if_app_3_in_1_gaa₅(x1, x2, x3, x4, x5) = if_app_3_in_1_gaa₂(x2, x5)
if_perm_2_in_3_ag₅(x1, x2, x3, x4, x5) = if_perm_2_in_3_ag₂(x3, x5)
PERM_2_IN_AG₂(x1, x2) = PERM_2_IN_AG₁(x2)
IF_PERM_2_IN_2_AG₆(x1, x2, x3, x4, x5, x6) = IF_PERM_2_IN_2_AG₂(x3, x6)
IF_PERM_2_IN_1_AG₄(x1, x2, x3, x4) = IF_PERM_2_IN_1_AG₂(x3, x4)

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

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

PERM_2_IN_AG₂(Xs, ._2₂(X, Ys)) -> IF_PERM_2_IN_1_AG₄(Xs, X, Ys, app_3_in_aaa₃(X1s, ._2₂(X, X2s), Xs))
IF_PERM_2_IN_1_AG₄(Xs, X, Ys, app_3_out_aaa₃(X1s, ._2₂(X, X2s), Xs)) -> IF_PERM_2_IN_2_AG₆(Xs, X, Ys, X1s, X2s, app_3_in_gaa₃(X1s, X2s, Zs))
IF_PERM_2_IN_2_AG₆(Xs, X, Ys, X1s, X2s, app_3_out_gaa₃(X1s, X2s, Zs)) -> PERM_2_IN_AG₂(Zs, Ys)

The TRS R consists of the following rules:

app_3_in_aaa₃([]_0, X, X) -> app_3_out_aaa₃([]_0, X, X)
app_3_in_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_in_aaa₃(Xs, Ys, Zs))
app_3_in_gaa₃([]_0, X, X) -> app_3_out_gaa₃([]_0, X, X)
app_3_in_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_in_gaa₃(Xs, Ys, Zs))
if_app_3_in_1_aaa₅(X, Xs, Ys, Zs, app_3_out_aaa₃(Xs, Ys, Zs)) -> app_3_out_aaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_app_3_in_1_gaa₅(X, Xs, Ys, Zs, app_3_out_gaa₃(Xs, Ys, Zs)) -> app_3_out_gaa₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))

The argument filtering Pi contains the following mapping:
[]_0 = []_0
._2₂(x1, x2) = ._2₁(x2)
app_3_in_aaa₃(x1, x2, x3) = app_3_in_aaa
app_3_out_aaa₃(x1, x2, x3) = app_3_out_aaa₁(x1)
if_app_3_in_1_aaa₅(x1, x2, x3, x4, x5) = if_app_3_in_1_aaa₁(x5)
app_3_in_gaa₃(x1, x2, x3) = app_3_in_gaa₁(x1)
app_3_out_gaa₃(x1, x2, x3) = app_3_out_gaa₁(x1)
if_app_3_in_1_gaa₅(x1, x2, x3, x4, x5) = if_app_3_in_1_gaa₂(x2, x5)
PERM_2_IN_AG₂(x1, x2) = PERM_2_IN_AG₁(x2)
IF_PERM_2_IN_2_AG₆(x1, x2, x3, x4, x5, x6) = IF_PERM_2_IN_2_AG₂(x3, x6)
IF_PERM_2_IN_1_AG₄(x1, x2, x3, x4) = IF_PERM_2_IN_1_AG₂(x3, x4)

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

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

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

PERM_2_IN_AG₁(._2₁(Ys)) -> IF_PERM_2_IN_1_AG₂(Ys, app_3_in_aaa)
IF_PERM_2_IN_1_AG₂(Ys, app_3_out_aaa₁(X1s)) -> IF_PERM_2_IN_2_AG₂(Ys, app_3_in_gaa₁(X1s))
IF_PERM_2_IN_2_AG₂(Ys, app_3_out_gaa₁(X1s)) -> PERM_2_IN_AG₁(Ys)

The TRS R consists of the following rules:

app_3_in_aaa -> app_3_out_aaa₁([]_0)
app_3_in_aaa -> if_app_3_in_1_aaa₁(app_3_in_aaa)
app_3_in_gaa₁([]_0) -> app_3_out_gaa₁([]_0)
app_3_in_gaa₁(._2₁(Xs)) -> if_app_3_in_1_gaa₂(Xs, app_3_in_gaa₁(Xs))
if_app_3_in_1_aaa₁(app_3_out_aaa₁(Xs)) -> app_3_out_aaa₁(._2₁(Xs))
if_app_3_in_1_gaa₂(Xs, app_3_out_gaa₁(Xs)) -> app_3_out_gaa₁(._2₁(Xs))

The set Q consists of the following terms:

app_3_in_aaa
app_3_in_gaa₁(x0)
if_app_3_in_1_aaa₁(x0)
if_app_3_in_1_gaa₂(x0, x1)

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

IF_PERM_2_IN_1_AG₂(Ys, app_3_out_aaa₁(X1s)) -> IF_PERM_2_IN_2_AG₂(Ys, app_3_in_gaa₁(X1s))
The graph contains the following edges 1 >= 1
IF_PERM_2_IN_2_AG₂(Ys, app_3_out_gaa₁(X1s)) -> PERM_2_IN_AG₁(Ys)
The graph contains the following edges 1 >= 1
PERM_2_IN_AG₁(._2₁(Ys)) -> IF_PERM_2_IN_1_AG₂(Ys, app_3_in_aaa)
The graph contains the following edges 1 > 1