Left Termination proof of /tmp/tmpJ40MRK/rev.pl

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

↳ PROLOG

  ↳ PrologToPiTRSProof

rev₂({}₀, {}₀).
rev₂(.₂(X, XS), .₂(Y, YS)) :- rev1₃(X, XS, Y), rev2₃(X, XS, YS).
rev1₃(X, {}₀, X).
rev1₃(X, .₂(Y, YS), Z) :- rev1₃(Y, YS, Z).
rev2₃(X, {}₀, {}₀).
rev2₃(X, .₂(Y, YS), ZS) :- rev2₃(Y, YS, US), rev₂(US, VS), rev₂(.₂(X, VS), ZS).

With regard to the inferred argument filtering the predicates were used in the following modes:
rev₂: (b,f)
rev1₃: (b,b,f)
rev2₃: (b,b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

rev_2_in_ga₂([]_0, []_0) -> rev_2_out_ga₂([]_0, []_0)
rev_2_in_ga₂(._2₂(X, XS), ._2₂(Y, YS)) -> if_rev_2_in_1_ga₅(X, XS, Y, YS, rev1_3_in_gga₃(X, XS, Y))
rev1_3_in_gga₃(X, []_0, X) -> rev1_3_out_gga₃(X, []_0, X)
rev1_3_in_gga₃(X, ._2₂(Y, YS), Z) -> if_rev1_3_in_1_gga₅(X, Y, YS, Z, rev1_3_in_gga₃(Y, YS, Z))
if_rev1_3_in_1_gga₅(X, Y, YS, Z, rev1_3_out_gga₃(Y, YS, Z)) -> rev1_3_out_gga₃(X, ._2₂(Y, YS), Z)
if_rev_2_in_1_ga₅(X, XS, Y, YS, rev1_3_out_gga₃(X, XS, Y)) -> if_rev_2_in_2_ga₅(X, XS, Y, YS, rev2_3_in_gga₃(X, XS, YS))
rev2_3_in_gga₃(X, []_0, []_0) -> rev2_3_out_gga₃(X, []_0, []_0)
rev2_3_in_gga₃(X, ._2₂(Y, YS), ZS) -> if_rev2_3_in_1_gga₅(X, Y, YS, ZS, rev2_3_in_gga₃(Y, YS, US))
if_rev2_3_in_1_gga₅(X, Y, YS, ZS, rev2_3_out_gga₃(Y, YS, US)) -> if_rev2_3_in_2_gga₆(X, Y, YS, ZS, US, rev_2_in_ga₂(US, VS))
if_rev2_3_in_2_gga₆(X, Y, YS, ZS, US, rev_2_out_ga₂(US, VS)) -> if_rev2_3_in_3_gga₆(X, Y, YS, ZS, VS, rev_2_in_ga₂(._2₂(X, VS), ZS))
if_rev2_3_in_3_gga₆(X, Y, YS, ZS, VS, rev_2_out_ga₂(._2₂(X, VS), ZS)) -> rev2_3_out_gga₃(X, ._2₂(Y, YS), ZS)
if_rev_2_in_2_ga₅(X, XS, Y, YS, rev2_3_out_gga₃(X, XS, YS)) -> rev_2_out_ga₂(._2₂(X, XS), ._2₂(Y, 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:

rev_2_in_ga₂([]_0, []_0) -> rev_2_out_ga₂([]_0, []_0)
rev_2_in_ga₂(._2₂(X, XS), ._2₂(Y, YS)) -> if_rev_2_in_1_ga₅(X, XS, Y, YS, rev1_3_in_gga₃(X, XS, Y))
rev1_3_in_gga₃(X, []_0, X) -> rev1_3_out_gga₃(X, []_0, X)
rev1_3_in_gga₃(X, ._2₂(Y, YS), Z) -> if_rev1_3_in_1_gga₅(X, Y, YS, Z, rev1_3_in_gga₃(Y, YS, Z))
if_rev1_3_in_1_gga₅(X, Y, YS, Z, rev1_3_out_gga₃(Y, YS, Z)) -> rev1_3_out_gga₃(X, ._2₂(Y, YS), Z)
if_rev_2_in_1_ga₅(X, XS, Y, YS, rev1_3_out_gga₃(X, XS, Y)) -> if_rev_2_in_2_ga₅(X, XS, Y, YS, rev2_3_in_gga₃(X, XS, YS))
rev2_3_in_gga₃(X, []_0, []_0) -> rev2_3_out_gga₃(X, []_0, []_0)
rev2_3_in_gga₃(X, ._2₂(Y, YS), ZS) -> if_rev2_3_in_1_gga₅(X, Y, YS, ZS, rev2_3_in_gga₃(Y, YS, US))
if_rev2_3_in_1_gga₅(X, Y, YS, ZS, rev2_3_out_gga₃(Y, YS, US)) -> if_rev2_3_in_2_gga₆(X, Y, YS, ZS, US, rev_2_in_ga₂(US, VS))
if_rev2_3_in_2_gga₆(X, Y, YS, ZS, US, rev_2_out_ga₂(US, VS)) -> if_rev2_3_in_3_gga₆(X, Y, YS, ZS, VS, rev_2_in_ga₂(._2₂(X, VS), ZS))
if_rev2_3_in_3_gga₆(X, Y, YS, ZS, VS, rev_2_out_ga₂(._2₂(X, VS), ZS)) -> rev2_3_out_gga₃(X, ._2₂(Y, YS), ZS)
if_rev_2_in_2_ga₅(X, XS, Y, YS, rev2_3_out_gga₃(X, XS, YS)) -> rev_2_out_ga₂(._2₂(X, XS), ._2₂(Y, YS))

REV_2_IN_GA₂(._2₂(X, XS), ._2₂(Y, YS)) -> IF_REV_2_IN_1_GA₅(X, XS, Y, YS, rev1_3_in_gga₃(X, XS, Y))
REV_2_IN_GA₂(._2₂(X, XS), ._2₂(Y, YS)) -> REV1_3_IN_GGA₃(X, XS, Y)
REV1_3_IN_GGA₃(X, ._2₂(Y, YS), Z) -> IF_REV1_3_IN_1_GGA₅(X, Y, YS, Z, rev1_3_in_gga₃(Y, YS, Z))
REV1_3_IN_GGA₃(X, ._2₂(Y, YS), Z) -> REV1_3_IN_GGA₃(Y, YS, Z)
IF_REV_2_IN_1_GA₅(X, XS, Y, YS, rev1_3_out_gga₃(X, XS, Y)) -> IF_REV_2_IN_2_GA₅(X, XS, Y, YS, rev2_3_in_gga₃(X, XS, YS))
IF_REV_2_IN_1_GA₅(X, XS, Y, YS, rev1_3_out_gga₃(X, XS, Y)) -> REV2_3_IN_GGA₃(X, XS, YS)
REV2_3_IN_GGA₃(X, ._2₂(Y, YS), ZS) -> IF_REV2_3_IN_1_GGA₅(X, Y, YS, ZS, rev2_3_in_gga₃(Y, YS, US))
REV2_3_IN_GGA₃(X, ._2₂(Y, YS), ZS) -> REV2_3_IN_GGA₃(Y, YS, US)
IF_REV2_3_IN_1_GGA₅(X, Y, YS, ZS, rev2_3_out_gga₃(Y, YS, US)) -> IF_REV2_3_IN_2_GGA₆(X, Y, YS, ZS, US, rev_2_in_ga₂(US, VS))
IF_REV2_3_IN_1_GGA₅(X, Y, YS, ZS, rev2_3_out_gga₃(Y, YS, US)) -> REV_2_IN_GA₂(US, VS)
IF_REV2_3_IN_2_GGA₆(X, Y, YS, ZS, US, rev_2_out_ga₂(US, VS)) -> IF_REV2_3_IN_3_GGA₆(X, Y, YS, ZS, VS, rev_2_in_ga₂(._2₂(X, VS), ZS))
IF_REV2_3_IN_2_GGA₆(X, Y, YS, ZS, US, rev_2_out_ga₂(US, VS)) -> REV_2_IN_GA₂(._2₂(X, VS), ZS)

The TRS R consists of the following rules:

rev_2_in_ga₂([]_0, []_0) -> rev_2_out_ga₂([]_0, []_0)
rev_2_in_ga₂(._2₂(X, XS), ._2₂(Y, YS)) -> if_rev_2_in_1_ga₅(X, XS, Y, YS, rev1_3_in_gga₃(X, XS, Y))
rev1_3_in_gga₃(X, []_0, X) -> rev1_3_out_gga₃(X, []_0, X)
rev1_3_in_gga₃(X, ._2₂(Y, YS), Z) -> if_rev1_3_in_1_gga₅(X, Y, YS, Z, rev1_3_in_gga₃(Y, YS, Z))
if_rev1_3_in_1_gga₅(X, Y, YS, Z, rev1_3_out_gga₃(Y, YS, Z)) -> rev1_3_out_gga₃(X, ._2₂(Y, YS), Z)
if_rev_2_in_1_ga₅(X, XS, Y, YS, rev1_3_out_gga₃(X, XS, Y)) -> if_rev_2_in_2_ga₅(X, XS, Y, YS, rev2_3_in_gga₃(X, XS, YS))
rev2_3_in_gga₃(X, []_0, []_0) -> rev2_3_out_gga₃(X, []_0, []_0)
rev2_3_in_gga₃(X, ._2₂(Y, YS), ZS) -> if_rev2_3_in_1_gga₅(X, Y, YS, ZS, rev2_3_in_gga₃(Y, YS, US))
if_rev2_3_in_1_gga₅(X, Y, YS, ZS, rev2_3_out_gga₃(Y, YS, US)) -> if_rev2_3_in_2_gga₆(X, Y, YS, ZS, US, rev_2_in_ga₂(US, VS))
if_rev2_3_in_2_gga₆(X, Y, YS, ZS, US, rev_2_out_ga₂(US, VS)) -> if_rev2_3_in_3_gga₆(X, Y, YS, ZS, VS, rev_2_in_ga₂(._2₂(X, VS), ZS))
if_rev2_3_in_3_gga₆(X, Y, YS, ZS, VS, rev_2_out_ga₂(._2₂(X, VS), ZS)) -> rev2_3_out_gga₃(X, ._2₂(Y, YS), ZS)
if_rev_2_in_2_ga₅(X, XS, Y, YS, rev2_3_out_gga₃(X, XS, YS)) -> rev_2_out_ga₂(._2₂(X, XS), ._2₂(Y, YS))

The argument filtering Pi contains the following mapping:
rev_2_in_ga₂(x1, x2) = rev_2_in_ga₁(x1)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
rev_2_out_ga₂(x1, x2) = rev_2_out_ga₁(x2)
if_rev_2_in_1_ga₅(x1, x2, x3, x4, x5) = if_rev_2_in_1_ga₃(x1, x2, x5)
rev1_3_in_gga₃(x1, x2, x3) = rev1_3_in_gga₂(x1, x2)
rev1_3_out_gga₃(x1, x2, x3) = rev1_3_out_gga₁(x3)
if_rev1_3_in_1_gga₅(x1, x2, x3, x4, x5) = if_rev1_3_in_1_gga₁(x5)
if_rev_2_in_2_ga₅(x1, x2, x3, x4, x5) = if_rev_2_in_2_ga₂(x3, x5)
rev2_3_in_gga₃(x1, x2, x3) = rev2_3_in_gga₂(x1, x2)
rev2_3_out_gga₃(x1, x2, x3) = rev2_3_out_gga₁(x3)
if_rev2_3_in_1_gga₅(x1, x2, x3, x4, x5) = if_rev2_3_in_1_gga₂(x1, x5)
if_rev2_3_in_2_gga₆(x1, x2, x3, x4, x5, x6) = if_rev2_3_in_2_gga₂(x1, x6)
if_rev2_3_in_3_gga₆(x1, x2, x3, x4, x5, x6) = if_rev2_3_in_3_gga₁(x6)
IF_REV_2_IN_2_GA₅(x1, x2, x3, x4, x5) = IF_REV_2_IN_2_GA₂(x3, x5)
IF_REV2_3_IN_1_GGA₅(x1, x2, x3, x4, x5) = IF_REV2_3_IN_1_GGA₂(x1, x5)
IF_REV1_3_IN_1_GGA₅(x1, x2, x3, x4, x5) = IF_REV1_3_IN_1_GGA₁(x5)
REV_2_IN_GA₂(x1, x2) = REV_2_IN_GA₁(x1)
IF_REV_2_IN_1_GA₅(x1, x2, x3, x4, x5) = IF_REV_2_IN_1_GA₃(x1, x2, x5)
IF_REV2_3_IN_3_GGA₆(x1, x2, x3, x4, x5, x6) = IF_REV2_3_IN_3_GGA₁(x6)
IF_REV2_3_IN_2_GGA₆(x1, x2, x3, x4, x5, x6) = IF_REV2_3_IN_2_GGA₂(x1, x6)
REV2_3_IN_GGA₃(x1, x2, x3) = REV2_3_IN_GGA₂(x1, x2)
REV1_3_IN_GGA₃(x1, x2, x3) = REV1_3_IN_GGA₂(x1, x2)

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:

REV_2_IN_GA₂(._2₂(X, XS), ._2₂(Y, YS)) -> IF_REV_2_IN_1_GA₅(X, XS, Y, YS, rev1_3_in_gga₃(X, XS, Y))
REV_2_IN_GA₂(._2₂(X, XS), ._2₂(Y, YS)) -> REV1_3_IN_GGA₃(X, XS, Y)
REV1_3_IN_GGA₃(X, ._2₂(Y, YS), Z) -> IF_REV1_3_IN_1_GGA₅(X, Y, YS, Z, rev1_3_in_gga₃(Y, YS, Z))
REV1_3_IN_GGA₃(X, ._2₂(Y, YS), Z) -> REV1_3_IN_GGA₃(Y, YS, Z)
IF_REV_2_IN_1_GA₅(X, XS, Y, YS, rev1_3_out_gga₃(X, XS, Y)) -> IF_REV_2_IN_2_GA₅(X, XS, Y, YS, rev2_3_in_gga₃(X, XS, YS))
IF_REV_2_IN_1_GA₅(X, XS, Y, YS, rev1_3_out_gga₃(X, XS, Y)) -> REV2_3_IN_GGA₃(X, XS, YS)
REV2_3_IN_GGA₃(X, ._2₂(Y, YS), ZS) -> IF_REV2_3_IN_1_GGA₅(X, Y, YS, ZS, rev2_3_in_gga₃(Y, YS, US))
REV2_3_IN_GGA₃(X, ._2₂(Y, YS), ZS) -> REV2_3_IN_GGA₃(Y, YS, US)
IF_REV2_3_IN_1_GGA₅(X, Y, YS, ZS, rev2_3_out_gga₃(Y, YS, US)) -> IF_REV2_3_IN_2_GGA₆(X, Y, YS, ZS, US, rev_2_in_ga₂(US, VS))
IF_REV2_3_IN_1_GGA₅(X, Y, YS, ZS, rev2_3_out_gga₃(Y, YS, US)) -> REV_2_IN_GA₂(US, VS)
IF_REV2_3_IN_2_GGA₆(X, Y, YS, ZS, US, rev_2_out_ga₂(US, VS)) -> IF_REV2_3_IN_3_GGA₆(X, Y, YS, ZS, VS, rev_2_in_ga₂(._2₂(X, VS), ZS))
IF_REV2_3_IN_2_GGA₆(X, Y, YS, ZS, US, rev_2_out_ga₂(US, VS)) -> REV_2_IN_GA₂(._2₂(X, VS), ZS)

The TRS R consists of the following rules:

rev_2_in_ga₂([]_0, []_0) -> rev_2_out_ga₂([]_0, []_0)
rev_2_in_ga₂(._2₂(X, XS), ._2₂(Y, YS)) -> if_rev_2_in_1_ga₅(X, XS, Y, YS, rev1_3_in_gga₃(X, XS, Y))
rev1_3_in_gga₃(X, []_0, X) -> rev1_3_out_gga₃(X, []_0, X)
rev1_3_in_gga₃(X, ._2₂(Y, YS), Z) -> if_rev1_3_in_1_gga₅(X, Y, YS, Z, rev1_3_in_gga₃(Y, YS, Z))
if_rev1_3_in_1_gga₅(X, Y, YS, Z, rev1_3_out_gga₃(Y, YS, Z)) -> rev1_3_out_gga₃(X, ._2₂(Y, YS), Z)
if_rev_2_in_1_ga₅(X, XS, Y, YS, rev1_3_out_gga₃(X, XS, Y)) -> if_rev_2_in_2_ga₅(X, XS, Y, YS, rev2_3_in_gga₃(X, XS, YS))
rev2_3_in_gga₃(X, []_0, []_0) -> rev2_3_out_gga₃(X, []_0, []_0)
rev2_3_in_gga₃(X, ._2₂(Y, YS), ZS) -> if_rev2_3_in_1_gga₅(X, Y, YS, ZS, rev2_3_in_gga₃(Y, YS, US))
if_rev2_3_in_1_gga₅(X, Y, YS, ZS, rev2_3_out_gga₃(Y, YS, US)) -> if_rev2_3_in_2_gga₆(X, Y, YS, ZS, US, rev_2_in_ga₂(US, VS))
if_rev2_3_in_2_gga₆(X, Y, YS, ZS, US, rev_2_out_ga₂(US, VS)) -> if_rev2_3_in_3_gga₆(X, Y, YS, ZS, VS, rev_2_in_ga₂(._2₂(X, VS), ZS))
if_rev2_3_in_3_gga₆(X, Y, YS, ZS, VS, rev_2_out_ga₂(._2₂(X, VS), ZS)) -> rev2_3_out_gga₃(X, ._2₂(Y, YS), ZS)
if_rev_2_in_2_ga₅(X, XS, Y, YS, rev2_3_out_gga₃(X, XS, YS)) -> rev_2_out_ga₂(._2₂(X, XS), ._2₂(Y, YS))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

REV1_3_IN_GGA₃(X, ._2₂(Y, YS), Z) -> REV1_3_IN_GGA₃(Y, YS, Z)

The TRS R consists of the following rules:

rev_2_in_ga₂([]_0, []_0) -> rev_2_out_ga₂([]_0, []_0)
rev_2_in_ga₂(._2₂(X, XS), ._2₂(Y, YS)) -> if_rev_2_in_1_ga₅(X, XS, Y, YS, rev1_3_in_gga₃(X, XS, Y))
rev1_3_in_gga₃(X, []_0, X) -> rev1_3_out_gga₃(X, []_0, X)
rev1_3_in_gga₃(X, ._2₂(Y, YS), Z) -> if_rev1_3_in_1_gga₅(X, Y, YS, Z, rev1_3_in_gga₃(Y, YS, Z))
if_rev1_3_in_1_gga₅(X, Y, YS, Z, rev1_3_out_gga₃(Y, YS, Z)) -> rev1_3_out_gga₃(X, ._2₂(Y, YS), Z)
if_rev_2_in_1_ga₅(X, XS, Y, YS, rev1_3_out_gga₃(X, XS, Y)) -> if_rev_2_in_2_ga₅(X, XS, Y, YS, rev2_3_in_gga₃(X, XS, YS))
rev2_3_in_gga₃(X, []_0, []_0) -> rev2_3_out_gga₃(X, []_0, []_0)
rev2_3_in_gga₃(X, ._2₂(Y, YS), ZS) -> if_rev2_3_in_1_gga₅(X, Y, YS, ZS, rev2_3_in_gga₃(Y, YS, US))
if_rev2_3_in_1_gga₅(X, Y, YS, ZS, rev2_3_out_gga₃(Y, YS, US)) -> if_rev2_3_in_2_gga₆(X, Y, YS, ZS, US, rev_2_in_ga₂(US, VS))
if_rev2_3_in_2_gga₆(X, Y, YS, ZS, US, rev_2_out_ga₂(US, VS)) -> if_rev2_3_in_3_gga₆(X, Y, YS, ZS, VS, rev_2_in_ga₂(._2₂(X, VS), ZS))
if_rev2_3_in_3_gga₆(X, Y, YS, ZS, VS, rev_2_out_ga₂(._2₂(X, VS), ZS)) -> rev2_3_out_gga₃(X, ._2₂(Y, YS), ZS)
if_rev_2_in_2_ga₅(X, XS, Y, YS, rev2_3_out_gga₃(X, XS, YS)) -> rev_2_out_ga₂(._2₂(X, XS), ._2₂(Y, YS))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

REV1_3_IN_GGA₃(X, ._2₂(Y, YS), Z) -> REV1_3_IN_GGA₃(Y, YS, Z)

R is empty.
The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
REV1_3_IN_GGA₃(x1, x2, x3) = REV1_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

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

REV1_3_IN_GGA₂(X, ._2₂(Y, YS)) -> REV1_3_IN_GGA₂(Y, YS)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {REV1_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:

REV1_3_IN_GGA₂(X, ._2₂(Y, YS)) -> REV1_3_IN_GGA₂(Y, YS)
The graph contains the following edges 2 > 1, 2 > 2

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

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

IF_REV2_3_IN_1_GGA₅(X, Y, YS, ZS, rev2_3_out_gga₃(Y, YS, US)) -> REV_2_IN_GA₂(US, VS)
REV2_3_IN_GGA₃(X, ._2₂(Y, YS), ZS) -> REV2_3_IN_GGA₃(Y, YS, US)
IF_REV2_3_IN_1_GGA₅(X, Y, YS, ZS, rev2_3_out_gga₃(Y, YS, US)) -> IF_REV2_3_IN_2_GGA₆(X, Y, YS, ZS, US, rev_2_in_ga₂(US, VS))
IF_REV_2_IN_1_GA₅(X, XS, Y, YS, rev1_3_out_gga₃(X, XS, Y)) -> REV2_3_IN_GGA₃(X, XS, YS)
REV2_3_IN_GGA₃(X, ._2₂(Y, YS), ZS) -> IF_REV2_3_IN_1_GGA₅(X, Y, YS, ZS, rev2_3_in_gga₃(Y, YS, US))
REV_2_IN_GA₂(._2₂(X, XS), ._2₂(Y, YS)) -> IF_REV_2_IN_1_GA₅(X, XS, Y, YS, rev1_3_in_gga₃(X, XS, Y))
IF_REV2_3_IN_2_GGA₆(X, Y, YS, ZS, US, rev_2_out_ga₂(US, VS)) -> REV_2_IN_GA₂(._2₂(X, VS), ZS)

The TRS R consists of the following rules:

rev_2_in_ga₂([]_0, []_0) -> rev_2_out_ga₂([]_0, []_0)
rev_2_in_ga₂(._2₂(X, XS), ._2₂(Y, YS)) -> if_rev_2_in_1_ga₅(X, XS, Y, YS, rev1_3_in_gga₃(X, XS, Y))
rev1_3_in_gga₃(X, []_0, X) -> rev1_3_out_gga₃(X, []_0, X)
rev1_3_in_gga₃(X, ._2₂(Y, YS), Z) -> if_rev1_3_in_1_gga₅(X, Y, YS, Z, rev1_3_in_gga₃(Y, YS, Z))
if_rev1_3_in_1_gga₅(X, Y, YS, Z, rev1_3_out_gga₃(Y, YS, Z)) -> rev1_3_out_gga₃(X, ._2₂(Y, YS), Z)
if_rev_2_in_1_ga₅(X, XS, Y, YS, rev1_3_out_gga₃(X, XS, Y)) -> if_rev_2_in_2_ga₅(X, XS, Y, YS, rev2_3_in_gga₃(X, XS, YS))
rev2_3_in_gga₃(X, []_0, []_0) -> rev2_3_out_gga₃(X, []_0, []_0)
rev2_3_in_gga₃(X, ._2₂(Y, YS), ZS) -> if_rev2_3_in_1_gga₅(X, Y, YS, ZS, rev2_3_in_gga₃(Y, YS, US))
if_rev2_3_in_1_gga₅(X, Y, YS, ZS, rev2_3_out_gga₃(Y, YS, US)) -> if_rev2_3_in_2_gga₆(X, Y, YS, ZS, US, rev_2_in_ga₂(US, VS))
if_rev2_3_in_2_gga₆(X, Y, YS, ZS, US, rev_2_out_ga₂(US, VS)) -> if_rev2_3_in_3_gga₆(X, Y, YS, ZS, VS, rev_2_in_ga₂(._2₂(X, VS), ZS))
if_rev2_3_in_3_gga₆(X, Y, YS, ZS, VS, rev_2_out_ga₂(._2₂(X, VS), ZS)) -> rev2_3_out_gga₃(X, ._2₂(Y, YS), ZS)
if_rev_2_in_2_ga₅(X, XS, Y, YS, rev2_3_out_gga₃(X, XS, YS)) -> rev_2_out_ga₂(._2₂(X, XS), ._2₂(Y, YS))

The argument filtering Pi contains the following mapping:
rev_2_in_ga₂(x1, x2) = rev_2_in_ga₁(x1)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
rev_2_out_ga₂(x1, x2) = rev_2_out_ga₁(x2)
if_rev_2_in_1_ga₅(x1, x2, x3, x4, x5) = if_rev_2_in_1_ga₃(x1, x2, x5)
rev1_3_in_gga₃(x1, x2, x3) = rev1_3_in_gga₂(x1, x2)
rev1_3_out_gga₃(x1, x2, x3) = rev1_3_out_gga₁(x3)
if_rev1_3_in_1_gga₅(x1, x2, x3, x4, x5) = if_rev1_3_in_1_gga₁(x5)
if_rev_2_in_2_ga₅(x1, x2, x3, x4, x5) = if_rev_2_in_2_ga₂(x3, x5)
rev2_3_in_gga₃(x1, x2, x3) = rev2_3_in_gga₂(x1, x2)
rev2_3_out_gga₃(x1, x2, x3) = rev2_3_out_gga₁(x3)
if_rev2_3_in_1_gga₅(x1, x2, x3, x4, x5) = if_rev2_3_in_1_gga₂(x1, x5)
if_rev2_3_in_2_gga₆(x1, x2, x3, x4, x5, x6) = if_rev2_3_in_2_gga₂(x1, x6)
if_rev2_3_in_3_gga₆(x1, x2, x3, x4, x5, x6) = if_rev2_3_in_3_gga₁(x6)
IF_REV2_3_IN_1_GGA₅(x1, x2, x3, x4, x5) = IF_REV2_3_IN_1_GGA₂(x1, x5)
REV_2_IN_GA₂(x1, x2) = REV_2_IN_GA₁(x1)
IF_REV_2_IN_1_GA₅(x1, x2, x3, x4, x5) = IF_REV_2_IN_1_GA₃(x1, x2, x5)
IF_REV2_3_IN_2_GGA₆(x1, x2, x3, x4, x5, x6) = IF_REV2_3_IN_2_GGA₂(x1, x6)
REV2_3_IN_GGA₃(x1, x2, x3) = REV2_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

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPPoloProof

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

IF_REV2_3_IN_1_GGA₂(X, rev2_3_out_gga₁(US)) -> REV_2_IN_GA₁(US)
REV2_3_IN_GGA₂(X, ._2₂(Y, YS)) -> REV2_3_IN_GGA₂(Y, YS)
IF_REV2_3_IN_1_GGA₂(X, rev2_3_out_gga₁(US)) -> IF_REV2_3_IN_2_GGA₂(X, rev_2_in_ga₁(US))
IF_REV_2_IN_1_GA₃(X, XS, rev1_3_out_gga₁(Y)) -> REV2_3_IN_GGA₂(X, XS)
REV2_3_IN_GGA₂(X, ._2₂(Y, YS)) -> IF_REV2_3_IN_1_GGA₂(X, rev2_3_in_gga₂(Y, YS))
REV_2_IN_GA₁(._2₂(X, XS)) -> IF_REV_2_IN_1_GA₃(X, XS, rev1_3_in_gga₂(X, XS))
IF_REV2_3_IN_2_GGA₂(X, rev_2_out_ga₁(VS)) -> REV_2_IN_GA₁(._2₂(X, VS))

The TRS R consists of the following rules:

rev_2_in_ga₁([]_0) -> rev_2_out_ga₁([]_0)
rev_2_in_ga₁(._2₂(X, XS)) -> if_rev_2_in_1_ga₃(X, XS, rev1_3_in_gga₂(X, XS))
rev1_3_in_gga₂(X, []_0) -> rev1_3_out_gga₁(X)
rev1_3_in_gga₂(X, ._2₂(Y, YS)) -> if_rev1_3_in_1_gga₁(rev1_3_in_gga₂(Y, YS))
if_rev1_3_in_1_gga₁(rev1_3_out_gga₁(Z)) -> rev1_3_out_gga₁(Z)
if_rev_2_in_1_ga₃(X, XS, rev1_3_out_gga₁(Y)) -> if_rev_2_in_2_ga₂(Y, rev2_3_in_gga₂(X, XS))
rev2_3_in_gga₂(X, []_0) -> rev2_3_out_gga₁([]_0)
rev2_3_in_gga₂(X, ._2₂(Y, YS)) -> if_rev2_3_in_1_gga₂(X, rev2_3_in_gga₂(Y, YS))
if_rev2_3_in_1_gga₂(X, rev2_3_out_gga₁(US)) -> if_rev2_3_in_2_gga₂(X, rev_2_in_ga₁(US))
if_rev2_3_in_2_gga₂(X, rev_2_out_ga₁(VS)) -> if_rev2_3_in_3_gga₁(rev_2_in_ga₁(._2₂(X, VS)))
if_rev2_3_in_3_gga₁(rev_2_out_ga₁(ZS)) -> rev2_3_out_gga₁(ZS)
if_rev_2_in_2_ga₂(Y, rev2_3_out_gga₁(YS)) -> rev_2_out_ga₁(._2₂(Y, YS))

The set Q consists of the following terms:

rev_2_in_ga₁(x0)
rev1_3_in_gga₂(x0, x1)
if_rev1_3_in_1_gga₁(x0)
if_rev_2_in_1_ga₃(x0, x1, x2)
rev2_3_in_gga₂(x0, x1)
if_rev2_3_in_1_gga₂(x0, x1)
if_rev2_3_in_2_gga₂(x0, x1)
if_rev2_3_in_3_gga₁(x0)
if_rev_2_in_2_ga₂(x0, x1)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {REV_2_IN_GA₁, IF_REV2_3_IN_1_GGA₂, REV2_3_IN_GGA₂, IF_REV2_3_IN_2_GGA₂, IF_REV_2_IN_1_GA₃}.
By using a polynomial ordering, the following set of Dependency Pairs of this DP problem can be strictly oriented.

IF_REV2_3_IN_1_GGA₂(X, rev2_3_out_gga₁(US)) -> REV_2_IN_GA₁(US)
REV2_3_IN_GGA₂(X, ._2₂(Y, YS)) -> REV2_3_IN_GGA₂(Y, YS)
REV_2_IN_GA₁(._2₂(X, XS)) -> IF_REV_2_IN_1_GA₃(X, XS, rev1_3_in_gga₂(X, XS))

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

IF_REV2_3_IN_1_GGA₂(X, rev2_3_out_gga₁(US)) -> IF_REV2_3_IN_2_GGA₂(X, rev_2_in_ga₁(US))
IF_REV_2_IN_1_GA₃(X, XS, rev1_3_out_gga₁(Y)) -> REV2_3_IN_GGA₂(X, XS)
REV2_3_IN_GGA₂(X, ._2₂(Y, YS)) -> IF_REV2_3_IN_1_GGA₂(X, rev2_3_in_gga₂(Y, YS))
IF_REV2_3_IN_2_GGA₂(X, rev_2_out_ga₁(VS)) -> REV_2_IN_GA₁(._2₂(X, VS))

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

rev2_3_in_gga₂(X, []_0) -> rev2_3_out_gga₁([]_0)
if_rev2_3_in_3_gga₁(rev_2_out_ga₁(ZS)) -> rev2_3_out_gga₁(ZS)
if_rev_2_in_1_ga₃(X, XS, rev1_3_out_gga₁(Y)) -> if_rev_2_in_2_ga₂(Y, rev2_3_in_gga₂(X, XS))
if_rev2_3_in_2_gga₂(X, rev_2_out_ga₁(VS)) -> if_rev2_3_in_3_gga₁(rev_2_in_ga₁(._2₂(X, VS)))
rev_2_in_ga₁([]_0) -> rev_2_out_ga₁([]_0)
if_rev2_3_in_1_gga₂(X, rev2_3_out_gga₁(US)) -> if_rev2_3_in_2_gga₂(X, rev_2_in_ga₁(US))
rev_2_in_ga₁(._2₂(X, XS)) -> if_rev_2_in_1_ga₃(X, XS, rev1_3_in_gga₂(X, XS))
rev2_3_in_gga₂(X, ._2₂(Y, YS)) -> if_rev2_3_in_1_gga₂(X, rev2_3_in_gga₂(Y, YS))
if_rev_2_in_2_ga₂(Y, rev2_3_out_gga₁(YS)) -> rev_2_out_ga₁(._2₂(Y, YS))

Used ordering: POLO with Polynomial interpretation:

POL(rev2_3_out_gga₁(x₁)) = x₁
POL(rev1_3_out_gga₁(x₁)) = 0
POL(if_rev2_3_in_2_gga₂(x₁, x₂)) = 1 + x₂
POL(if_rev2_3_in_1_gga₂(x₁, x₂)) = 1 + x₂
POL(REV_2_IN_GA₁(x₁)) = x₁
POL(if_rev2_3_in_3_gga₁(x₁)) = x₁
POL(rev2_3_in_gga₂(x₁, x₂)) = x₂
POL(rev1_3_in_gga₂(x₁, x₂)) = 0
POL([]_0) = 0
POL(if_rev1_3_in_1_gga₁(x₁)) = 0
POL(IF_REV_2_IN_1_GA₃(x₁, x₂, x₃)) = x₂
POL(REV2_3_IN_GGA₂(x₁, x₂)) = x₂
POL(rev_2_in_ga₁(x₁)) = x₁
POL(._2₂(x₁, x₂)) = 1 + x₂
POL(if_rev_2_in_1_ga₃(x₁, x₂, x₃)) = 1 + x₂
POL(rev_2_out_ga₁(x₁)) = x₁
POL(if_rev_2_in_2_ga₂(x₁, x₂)) = 1 + x₂
POL(IF_REV2_3_IN_1_GGA₂(x₁, x₂)) = 1 + x₂
POL(IF_REV2_3_IN_2_GGA₂(x₁, x₂)) = 1 + x₂

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPPoloProof

                      ↳ QDP

                        ↳ DependencyGraphProof

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

IF_REV2_3_IN_1_GGA₂(X, rev2_3_out_gga₁(US)) -> IF_REV2_3_IN_2_GGA₂(X, rev_2_in_ga₁(US))
IF_REV_2_IN_1_GA₃(X, XS, rev1_3_out_gga₁(Y)) -> REV2_3_IN_GGA₂(X, XS)
REV2_3_IN_GGA₂(X, ._2₂(Y, YS)) -> IF_REV2_3_IN_1_GGA₂(X, rev2_3_in_gga₂(Y, YS))
IF_REV2_3_IN_2_GGA₂(X, rev_2_out_ga₁(VS)) -> REV_2_IN_GA₁(._2₂(X, VS))

The TRS R consists of the following rules:

rev_2_in_ga₁([]_0) -> rev_2_out_ga₁([]_0)
rev_2_in_ga₁(._2₂(X, XS)) -> if_rev_2_in_1_ga₃(X, XS, rev1_3_in_gga₂(X, XS))
rev1_3_in_gga₂(X, []_0) -> rev1_3_out_gga₁(X)
rev1_3_in_gga₂(X, ._2₂(Y, YS)) -> if_rev1_3_in_1_gga₁(rev1_3_in_gga₂(Y, YS))
if_rev1_3_in_1_gga₁(rev1_3_out_gga₁(Z)) -> rev1_3_out_gga₁(Z)
if_rev_2_in_1_ga₃(X, XS, rev1_3_out_gga₁(Y)) -> if_rev_2_in_2_ga₂(Y, rev2_3_in_gga₂(X, XS))
rev2_3_in_gga₂(X, []_0) -> rev2_3_out_gga₁([]_0)
rev2_3_in_gga₂(X, ._2₂(Y, YS)) -> if_rev2_3_in_1_gga₂(X, rev2_3_in_gga₂(Y, YS))
if_rev2_3_in_1_gga₂(X, rev2_3_out_gga₁(US)) -> if_rev2_3_in_2_gga₂(X, rev_2_in_ga₁(US))
if_rev2_3_in_2_gga₂(X, rev_2_out_ga₁(VS)) -> if_rev2_3_in_3_gga₁(rev_2_in_ga₁(._2₂(X, VS)))
if_rev2_3_in_3_gga₁(rev_2_out_ga₁(ZS)) -> rev2_3_out_gga₁(ZS)
if_rev_2_in_2_ga₂(Y, rev2_3_out_gga₁(YS)) -> rev_2_out_ga₁(._2₂(Y, YS))

The set Q consists of the following terms:

rev_2_in_ga₁(x0)
rev1_3_in_gga₂(x0, x1)
if_rev1_3_in_1_gga₁(x0)
if_rev_2_in_1_ga₃(x0, x1, x2)
rev2_3_in_gga₂(x0, x1)
if_rev2_3_in_1_gga₂(x0, x1)
if_rev2_3_in_2_gga₂(x0, x1)
if_rev2_3_in_3_gga₁(x0)
if_rev_2_in_2_ga₂(x0, x1)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_REV2_3_IN_2_GGA₂, IF_REV2_3_IN_1_GGA₂, REV2_3_IN_GGA₂, IF_REV_2_IN_1_GA₃, REV_2_IN_GA₁}.
The approximation of the Dependency Graph contains 0 SCCs with 4 less nodes.