Left Termination proof of /tmp/tmp4AntJe/permutation-bf.pl

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

↳ PROLOG

  ↳ 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: (b,f)
app₃: (f,f,b) (b,b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_in_aag₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aag₃([]_0, X, X) -> app_3_out_aag₃([]_0, X, X)
app_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_in_aag₃(Xs, Ys, Zs))
if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_out_aag₃(Xs, Ys, Zs)) -> app_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_in_gga₃(X1s, X2s, Zs))
app_3_in_gga₃([]_0, X, X) -> app_3_out_gga₃([]_0, X, X)
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_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_out_ga₂(Zs, Ys)) -> perm_2_out_ga₂(Xs, ._2₂(X, Ys))

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

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

perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_in_aag₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aag₃([]_0, X, X) -> app_3_out_aag₃([]_0, X, X)
app_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_in_aag₃(Xs, Ys, Zs))
if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_out_aag₃(Xs, Ys, Zs)) -> app_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_in_gga₃(X1s, X2s, Zs))
app_3_in_gga₃([]_0, X, X) -> app_3_out_gga₃([]_0, X, X)
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_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_out_ga₂(Zs, Ys)) -> perm_2_out_ga₂(Xs, ._2₂(X, Ys))

PERM_2_IN_GA₂(Xs, ._2₂(X, Ys)) -> IF_PERM_2_IN_1_GA₄(Xs, X, Ys, app_3_in_aag₃(X1s, ._2₂(X, X2s), Xs))
PERM_2_IN_GA₂(Xs, ._2₂(X, Ys)) -> APP_3_IN_AAG₃(X1s, ._2₂(X, X2s), Xs)
APP_3_IN_AAG₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> IF_APP_3_IN_1_AAG₅(X, Xs, Ys, Zs, app_3_in_aag₃(Xs, Ys, Zs))
APP_3_IN_AAG₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_AAG₃(Xs, Ys, Zs)
IF_PERM_2_IN_1_GA₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, app_3_in_gga₃(X1s, X2s, Zs))
IF_PERM_2_IN_1_GA₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> APP_3_IN_GGA₃(X1s, X2s, Zs)
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_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> IF_PERM_2_IN_3_GA₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> PERM_2_IN_GA₂(Zs, Ys)

The TRS R consists of the following rules:

perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_in_aag₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aag₃([]_0, X, X) -> app_3_out_aag₃([]_0, X, X)
app_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_in_aag₃(Xs, Ys, Zs))
if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_out_aag₃(Xs, Ys, Zs)) -> app_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_in_gga₃(X1s, X2s, Zs))
app_3_in_gga₃([]_0, X, X) -> app_3_out_gga₃([]_0, X, X)
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_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_out_ga₂(Zs, Ys)) -> perm_2_out_ga₂(Xs, ._2₂(X, Ys))

The argument filtering Pi contains the following mapping:
perm_2_in_ga₂(x1, x2) = perm_2_in_ga₁(x1)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
perm_2_out_ga₂(x1, x2) = perm_2_out_ga₁(x2)
if_perm_2_in_1_ga₄(x1, x2, x3, x4) = if_perm_2_in_1_ga₁(x4)
app_3_in_aag₃(x1, x2, x3) = app_3_in_aag₁(x3)
app_3_out_aag₃(x1, x2, x3) = app_3_out_aag₂(x1, x2)
if_app_3_in_1_aag₅(x1, x2, x3, x4, x5) = if_app_3_in_1_aag₂(x1, x5)
if_perm_2_in_2_ga₆(x1, x2, x3, x4, x5, x6) = if_perm_2_in_2_ga₂(x2, x6)
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_perm_2_in_3_ga₅(x1, x2, x3, x4, x5) = if_perm_2_in_3_ga₂(x2, x5)
IF_APP_3_IN_1_GGA₅(x1, x2, x3, x4, x5) = IF_APP_3_IN_1_GGA₂(x1, x5)
IF_APP_3_IN_1_AAG₅(x1, x2, x3, x4, x5) = IF_APP_3_IN_1_AAG₂(x1, x5)
IF_PERM_2_IN_3_GA₅(x1, x2, x3, x4, x5) = IF_PERM_2_IN_3_GA₂(x2, x5)
APP_3_IN_GGA₃(x1, x2, x3) = APP_3_IN_GGA₂(x1, x2)
APP_3_IN_AAG₃(x1, x2, x3) = APP_3_IN_AAG₁(x3)
PERM_2_IN_GA₂(x1, x2) = PERM_2_IN_GA₁(x1)
IF_PERM_2_IN_2_GA₆(x1, x2, x3, x4, x5, x6) = IF_PERM_2_IN_2_GA₂(x2, x6)
IF_PERM_2_IN_1_GA₄(x1, x2, x3, x4) = IF_PERM_2_IN_1_GA₁(x4)

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

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

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

PERM_2_IN_GA₂(Xs, ._2₂(X, Ys)) -> IF_PERM_2_IN_1_GA₄(Xs, X, Ys, app_3_in_aag₃(X1s, ._2₂(X, X2s), Xs))
PERM_2_IN_GA₂(Xs, ._2₂(X, Ys)) -> APP_3_IN_AAG₃(X1s, ._2₂(X, X2s), Xs)
APP_3_IN_AAG₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> IF_APP_3_IN_1_AAG₅(X, Xs, Ys, Zs, app_3_in_aag₃(Xs, Ys, Zs))
APP_3_IN_AAG₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_AAG₃(Xs, Ys, Zs)
IF_PERM_2_IN_1_GA₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, app_3_in_gga₃(X1s, X2s, Zs))
IF_PERM_2_IN_1_GA₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> APP_3_IN_GGA₃(X1s, X2s, Zs)
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_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> IF_PERM_2_IN_3_GA₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> PERM_2_IN_GA₂(Zs, Ys)

The TRS R consists of the following rules:

perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_in_aag₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aag₃([]_0, X, X) -> app_3_out_aag₃([]_0, X, X)
app_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_in_aag₃(Xs, Ys, Zs))
if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_out_aag₃(Xs, Ys, Zs)) -> app_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_in_gga₃(X1s, X2s, Zs))
app_3_in_gga₃([]_0, X, X) -> app_3_out_gga₃([]_0, X, X)
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_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_out_ga₂(Zs, Ys)) -> perm_2_out_ga₂(Xs, ._2₂(X, Ys))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

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:

perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_in_aag₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aag₃([]_0, X, X) -> app_3_out_aag₃([]_0, X, X)
app_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_in_aag₃(Xs, Ys, Zs))
if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_out_aag₃(Xs, Ys, Zs)) -> app_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_in_gga₃(X1s, X2s, Zs))
app_3_in_gga₃([]_0, X, X) -> app_3_out_gga₃([]_0, X, X)
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_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_out_ga₂(Zs, Ys)) -> perm_2_out_ga₂(Xs, ._2₂(X, Ys))

↳ PROLOG

  ↳ 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_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

  ↳ 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_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

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

APP_3_IN_AAG₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_AAG₃(Xs, Ys, Zs)

The TRS R consists of the following rules:

perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_in_aag₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aag₃([]_0, X, X) -> app_3_out_aag₃([]_0, X, X)
app_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_in_aag₃(Xs, Ys, Zs))
if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_out_aag₃(Xs, Ys, Zs)) -> app_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_in_gga₃(X1s, X2s, Zs))
app_3_in_gga₃([]_0, X, X) -> app_3_out_gga₃([]_0, X, X)
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_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_out_ga₂(Zs, Ys)) -> perm_2_out_ga₂(Xs, ._2₂(X, Ys))

↳ PROLOG

  ↳ 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_AAG₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_AAG₃(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
APP_3_IN_AAG₃(x1, x2, x3) = APP_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

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

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

APP_3_IN_AAG₁(._2₂(X, Zs)) -> APP_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 {APP_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:

APP_3_IN_AAG₁(._2₂(X, Zs)) -> APP_3_IN_AAG₁(Zs)
The graph contains the following edges 1 > 1

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> PERM_2_IN_GA₂(Zs, Ys)
PERM_2_IN_GA₂(Xs, ._2₂(X, Ys)) -> IF_PERM_2_IN_1_GA₄(Xs, X, Ys, app_3_in_aag₃(X1s, ._2₂(X, X2s), Xs))
IF_PERM_2_IN_1_GA₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, app_3_in_gga₃(X1s, X2s, Zs))

The TRS R consists of the following rules:

perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_in_aag₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aag₃([]_0, X, X) -> app_3_out_aag₃([]_0, X, X)
app_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_in_aag₃(Xs, Ys, Zs))
if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_out_aag₃(Xs, Ys, Zs)) -> app_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_in_gga₃(X1s, X2s, Zs))
app_3_in_gga₃([]_0, X, X) -> app_3_out_gga₃([]_0, X, X)
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_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_out_ga₂(Zs, Ys)) -> perm_2_out_ga₂(Xs, ._2₂(X, Ys))

The argument filtering Pi contains the following mapping:
perm_2_in_ga₂(x1, x2) = perm_2_in_ga₁(x1)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
perm_2_out_ga₂(x1, x2) = perm_2_out_ga₁(x2)
if_perm_2_in_1_ga₄(x1, x2, x3, x4) = if_perm_2_in_1_ga₁(x4)
app_3_in_aag₃(x1, x2, x3) = app_3_in_aag₁(x3)
app_3_out_aag₃(x1, x2, x3) = app_3_out_aag₂(x1, x2)
if_app_3_in_1_aag₅(x1, x2, x3, x4, x5) = if_app_3_in_1_aag₂(x1, x5)
if_perm_2_in_2_ga₆(x1, x2, x3, x4, x5, x6) = if_perm_2_in_2_ga₂(x2, x6)
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_perm_2_in_3_ga₅(x1, x2, x3, x4, x5) = if_perm_2_in_3_ga₂(x2, x5)
PERM_2_IN_GA₂(x1, x2) = PERM_2_IN_GA₁(x1)
IF_PERM_2_IN_2_GA₆(x1, x2, x3, x4, x5, x6) = IF_PERM_2_IN_2_GA₂(x2, x6)
IF_PERM_2_IN_1_GA₄(x1, x2, x3, x4) = IF_PERM_2_IN_1_GA₁(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

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

IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> PERM_2_IN_GA₂(Zs, Ys)
PERM_2_IN_GA₂(Xs, ._2₂(X, Ys)) -> IF_PERM_2_IN_1_GA₄(Xs, X, Ys, app_3_in_aag₃(X1s, ._2₂(X, X2s), Xs))
IF_PERM_2_IN_1_GA₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, app_3_in_gga₃(X1s, X2s, Zs))

The TRS R consists of the following rules:

app_3_in_aag₃([]_0, X, X) -> app_3_out_aag₃([]_0, X, X)
app_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_in_aag₃(Xs, Ys, Zs))
app_3_in_gga₃([]_0, X, X) -> app_3_out_gga₃([]_0, X, X)
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_aag₅(X, Xs, Ys, Zs, app_3_out_aag₃(Xs, Ys, Zs)) -> app_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, 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))

The argument filtering Pi contains the following mapping:
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
app_3_in_aag₃(x1, x2, x3) = app_3_in_aag₁(x3)
app_3_out_aag₃(x1, x2, x3) = app_3_out_aag₂(x1, x2)
if_app_3_in_1_aag₅(x1, x2, x3, x4, x5) = if_app_3_in_1_aag₂(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)
PERM_2_IN_GA₂(x1, x2) = PERM_2_IN_GA₁(x1)
IF_PERM_2_IN_2_GA₆(x1, x2, x3, x4, x5, x6) = IF_PERM_2_IN_2_GA₂(x2, x6)
IF_PERM_2_IN_1_GA₄(x1, x2, x3, x4) = IF_PERM_2_IN_1_GA₁(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

                        ↳ QDPPoloProof

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

IF_PERM_2_IN_2_GA₂(X, app_3_out_gga₁(Zs)) -> PERM_2_IN_GA₁(Zs)
PERM_2_IN_GA₁(Xs) -> IF_PERM_2_IN_1_GA₁(app_3_in_aag₁(Xs))
IF_PERM_2_IN_1_GA₁(app_3_out_aag₂(X1s, ._2₂(X, X2s))) -> IF_PERM_2_IN_2_GA₂(X, app_3_in_gga₂(X1s, X2s))

The TRS R consists of the following rules:

app_3_in_aag₁(X) -> app_3_out_aag₂([]_0, X)
app_3_in_aag₁(._2₂(X, Zs)) -> if_app_3_in_1_aag₂(X, app_3_in_aag₁(Zs))
app_3_in_gga₂([]_0, X) -> app_3_out_gga₁(X)
app_3_in_gga₂(._2₂(X, Xs), Ys) -> if_app_3_in_1_gga₂(X, app_3_in_gga₂(Xs, Ys))
if_app_3_in_1_aag₂(X, app_3_out_aag₂(Xs, Ys)) -> app_3_out_aag₂(._2₂(X, Xs), Ys)
if_app_3_in_1_gga₂(X, app_3_out_gga₁(Zs)) -> app_3_out_gga₁(._2₂(X, Zs))

The set Q consists of the following terms:

app_3_in_aag₁(x0)
app_3_in_gga₂(x0, x1)
if_app_3_in_1_aag₂(x0, x1)
if_app_3_in_1_gga₂(x0, x1)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {PERM_2_IN_GA₁, IF_PERM_2_IN_2_GA₂, IF_PERM_2_IN_1_GA₁}.
By using a polynomial ordering, the following set of Dependency Pairs of this DP problem can be strictly oriented.

IF_PERM_2_IN_1_GA₁(app_3_out_aag₂(X1s, ._2₂(X, X2s))) -> IF_PERM_2_IN_2_GA₂(X, app_3_in_gga₂(X1s, X2s))

The remaining Dependency Pairs were at least non-strictly be oriented.

IF_PERM_2_IN_2_GA₂(X, app_3_out_gga₁(Zs)) -> PERM_2_IN_GA₁(Zs)
PERM_2_IN_GA₁(Xs) -> IF_PERM_2_IN_1_GA₁(app_3_in_aag₁(Xs))

With the implicit AFS we had to orient the following set of usable rules non-strictly.

app_3_in_gga₂(._2₂(X, Xs), Ys) -> if_app_3_in_1_gga₂(X, app_3_in_gga₂(Xs, Ys))
app_3_in_gga₂([]_0, X) -> app_3_out_gga₁(X)
app_3_in_aag₁(X) -> app_3_out_aag₂([]_0, X)
if_app_3_in_1_gga₂(X, app_3_out_gga₁(Zs)) -> app_3_out_gga₁(._2₂(X, Zs))
if_app_3_in_1_aag₂(X, app_3_out_aag₂(Xs, Ys)) -> app_3_out_aag₂(._2₂(X, Xs), Ys)
app_3_in_aag₁(._2₂(X, Zs)) -> if_app_3_in_1_aag₂(X, app_3_in_aag₁(Zs))

Used ordering: POLO with Polynomial interpretation:

POL(._2₂(x₁, x₂)) = 1 + x₂
POL(if_app_3_in_1_gga₂(x₁, x₂)) = 1 + x₂
POL(IF_PERM_2_IN_2_GA₂(x₁, x₂)) = x₂
POL(IF_PERM_2_IN_1_GA₁(x₁)) = x₁
POL(app_3_out_gga₁(x₁)) = x₁
POL(app_3_in_aag₁(x₁)) = x₁
POL(app_3_out_aag₂(x₁, x₂)) = x₁ + x₂
POL([]_0) = 0
POL(if_app_3_in_1_aag₂(x₁, x₂)) = 1 + x₂
POL(app_3_in_gga₂(x₁, x₂)) = x₁ + x₂
POL(PERM_2_IN_GA₁(x₁)) = x₁

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPPoloProof

                          ↳ QDP

                            ↳ DependencyGraphProof

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

IF_PERM_2_IN_2_GA₂(X, app_3_out_gga₁(Zs)) -> PERM_2_IN_GA₁(Zs)
PERM_2_IN_GA₁(Xs) -> IF_PERM_2_IN_1_GA₁(app_3_in_aag₁(Xs))

The TRS R consists of the following rules:

app_3_in_aag₁(X) -> app_3_out_aag₂([]_0, X)
app_3_in_aag₁(._2₂(X, Zs)) -> if_app_3_in_1_aag₂(X, app_3_in_aag₁(Zs))
app_3_in_gga₂([]_0, X) -> app_3_out_gga₁(X)
app_3_in_gga₂(._2₂(X, Xs), Ys) -> if_app_3_in_1_gga₂(X, app_3_in_gga₂(Xs, Ys))
if_app_3_in_1_aag₂(X, app_3_out_aag₂(Xs, Ys)) -> app_3_out_aag₂(._2₂(X, Xs), Ys)
if_app_3_in_1_gga₂(X, app_3_out_gga₁(Zs)) -> app_3_out_gga₁(._2₂(X, Zs))

The set Q consists of the following terms:

app_3_in_aag₁(x0)
app_3_in_gga₂(x0, x1)
if_app_3_in_1_aag₂(x0, x1)
if_app_3_in_1_gga₂(x0, x1)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {PERM_2_IN_GA₁, IF_PERM_2_IN_2_GA₂, IF_PERM_2_IN_1_GA₁}.
The approximation of the Dependency Graph contains 0 SCCs with 2 less nodes.