Left Termination proof of /tmp/tmpwXl8g2/samefringe.pl

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

↳ PROLOG

  ↳ PrologToPiTRSProof

gopher₂(nil₀, nil₀).
gopher₂(cons₂(nil₀, Y), cons₂(nil₀, Y)).
gopher₂(cons₂(cons₂(U, V), W), X) :- gopher₂(cons₂(U, cons₂(V, W)), X).
samefringe₂(nil₀, nil₀).
samefringe₂(cons₂(U, V), cons₂(X, Y)) :- gopher₂(cons₂(U, V), cons₂(U1, V1)), gopher₂(cons₂(X, Y), cons₂(X1, Y1)), samefringe₂(V1, Y1).

With regard to the inferred argument filtering the predicates were used in the following modes:
samefringe₂: (b,b)
gopher₂: (b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

samefringe_2_in_gg₂(nil_0, nil_0) -> samefringe_2_out_gg₂(nil_0, nil_0)
samefringe_2_in_gg₂(cons_2₂(U, V), cons_2₂(X, Y)) -> if_samefringe_2_in_1_gg₅(U, V, X, Y, gopher_2_in_ga₂(cons_2₂(U, V), cons_2₂(U1, V1)))
gopher_2_in_ga₂(nil_0, nil_0) -> gopher_2_out_ga₂(nil_0, nil_0)
gopher_2_in_ga₂(cons_2₂(nil_0, Y), cons_2₂(nil_0, Y)) -> gopher_2_out_ga₂(cons_2₂(nil_0, Y), cons_2₂(nil_0, Y))
gopher_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), X) -> if_gopher_2_in_1_ga₅(U, V, W, X, gopher_2_in_ga₂(cons_2₂(U, cons_2₂(V, W)), X))
if_gopher_2_in_1_ga₅(U, V, W, X, gopher_2_out_ga₂(cons_2₂(U, cons_2₂(V, W)), X)) -> gopher_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)
if_samefringe_2_in_1_gg₅(U, V, X, Y, gopher_2_out_ga₂(cons_2₂(U, V), cons_2₂(U1, V1))) -> if_samefringe_2_in_2_gg₆(U, V, X, Y, V1, gopher_2_in_ga₂(cons_2₂(X, Y), cons_2₂(X1, Y1)))
if_samefringe_2_in_2_gg₆(U, V, X, Y, V1, gopher_2_out_ga₂(cons_2₂(X, Y), cons_2₂(X1, Y1))) -> if_samefringe_2_in_3_gg₇(U, V, X, Y, V1, Y1, samefringe_2_in_gg₂(V1, Y1))
if_samefringe_2_in_3_gg₇(U, V, X, Y, V1, Y1, samefringe_2_out_gg₂(V1, Y1)) -> samefringe_2_out_gg₂(cons_2₂(U, V), cons_2₂(X, Y))

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:

samefringe_2_in_gg₂(nil_0, nil_0) -> samefringe_2_out_gg₂(nil_0, nil_0)
samefringe_2_in_gg₂(cons_2₂(U, V), cons_2₂(X, Y)) -> if_samefringe_2_in_1_gg₅(U, V, X, Y, gopher_2_in_ga₂(cons_2₂(U, V), cons_2₂(U1, V1)))
gopher_2_in_ga₂(nil_0, nil_0) -> gopher_2_out_ga₂(nil_0, nil_0)
gopher_2_in_ga₂(cons_2₂(nil_0, Y), cons_2₂(nil_0, Y)) -> gopher_2_out_ga₂(cons_2₂(nil_0, Y), cons_2₂(nil_0, Y))
gopher_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), X) -> if_gopher_2_in_1_ga₅(U, V, W, X, gopher_2_in_ga₂(cons_2₂(U, cons_2₂(V, W)), X))
if_gopher_2_in_1_ga₅(U, V, W, X, gopher_2_out_ga₂(cons_2₂(U, cons_2₂(V, W)), X)) -> gopher_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)
if_samefringe_2_in_1_gg₅(U, V, X, Y, gopher_2_out_ga₂(cons_2₂(U, V), cons_2₂(U1, V1))) -> if_samefringe_2_in_2_gg₆(U, V, X, Y, V1, gopher_2_in_ga₂(cons_2₂(X, Y), cons_2₂(X1, Y1)))
if_samefringe_2_in_2_gg₆(U, V, X, Y, V1, gopher_2_out_ga₂(cons_2₂(X, Y), cons_2₂(X1, Y1))) -> if_samefringe_2_in_3_gg₇(U, V, X, Y, V1, Y1, samefringe_2_in_gg₂(V1, Y1))
if_samefringe_2_in_3_gg₇(U, V, X, Y, V1, Y1, samefringe_2_out_gg₂(V1, Y1)) -> samefringe_2_out_gg₂(cons_2₂(U, V), cons_2₂(X, Y))

SAMEFRINGE_2_IN_GG₂(cons_2₂(U, V), cons_2₂(X, Y)) -> IF_SAMEFRINGE_2_IN_1_GG₅(U, V, X, Y, gopher_2_in_ga₂(cons_2₂(U, V), cons_2₂(U1, V1)))
SAMEFRINGE_2_IN_GG₂(cons_2₂(U, V), cons_2₂(X, Y)) -> GOPHER_2_IN_GA₂(cons_2₂(U, V), cons_2₂(U1, V1))
GOPHER_2_IN_GA₂(cons_2₂(cons_2₂(U, V), W), X) -> IF_GOPHER_2_IN_1_GA₅(U, V, W, X, gopher_2_in_ga₂(cons_2₂(U, cons_2₂(V, W)), X))
GOPHER_2_IN_GA₂(cons_2₂(cons_2₂(U, V), W), X) -> GOPHER_2_IN_GA₂(cons_2₂(U, cons_2₂(V, W)), X)
IF_SAMEFRINGE_2_IN_1_GG₅(U, V, X, Y, gopher_2_out_ga₂(cons_2₂(U, V), cons_2₂(U1, V1))) -> IF_SAMEFRINGE_2_IN_2_GG₆(U, V, X, Y, V1, gopher_2_in_ga₂(cons_2₂(X, Y), cons_2₂(X1, Y1)))
IF_SAMEFRINGE_2_IN_1_GG₅(U, V, X, Y, gopher_2_out_ga₂(cons_2₂(U, V), cons_2₂(U1, V1))) -> GOPHER_2_IN_GA₂(cons_2₂(X, Y), cons_2₂(X1, Y1))
IF_SAMEFRINGE_2_IN_2_GG₆(U, V, X, Y, V1, gopher_2_out_ga₂(cons_2₂(X, Y), cons_2₂(X1, Y1))) -> IF_SAMEFRINGE_2_IN_3_GG₇(U, V, X, Y, V1, Y1, samefringe_2_in_gg₂(V1, Y1))
IF_SAMEFRINGE_2_IN_2_GG₆(U, V, X, Y, V1, gopher_2_out_ga₂(cons_2₂(X, Y), cons_2₂(X1, Y1))) -> SAMEFRINGE_2_IN_GG₂(V1, Y1)

The TRS R consists of the following rules:

samefringe_2_in_gg₂(nil_0, nil_0) -> samefringe_2_out_gg₂(nil_0, nil_0)
samefringe_2_in_gg₂(cons_2₂(U, V), cons_2₂(X, Y)) -> if_samefringe_2_in_1_gg₅(U, V, X, Y, gopher_2_in_ga₂(cons_2₂(U, V), cons_2₂(U1, V1)))
gopher_2_in_ga₂(nil_0, nil_0) -> gopher_2_out_ga₂(nil_0, nil_0)
gopher_2_in_ga₂(cons_2₂(nil_0, Y), cons_2₂(nil_0, Y)) -> gopher_2_out_ga₂(cons_2₂(nil_0, Y), cons_2₂(nil_0, Y))
gopher_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), X) -> if_gopher_2_in_1_ga₅(U, V, W, X, gopher_2_in_ga₂(cons_2₂(U, cons_2₂(V, W)), X))
if_gopher_2_in_1_ga₅(U, V, W, X, gopher_2_out_ga₂(cons_2₂(U, cons_2₂(V, W)), X)) -> gopher_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)
if_samefringe_2_in_1_gg₅(U, V, X, Y, gopher_2_out_ga₂(cons_2₂(U, V), cons_2₂(U1, V1))) -> if_samefringe_2_in_2_gg₆(U, V, X, Y, V1, gopher_2_in_ga₂(cons_2₂(X, Y), cons_2₂(X1, Y1)))
if_samefringe_2_in_2_gg₆(U, V, X, Y, V1, gopher_2_out_ga₂(cons_2₂(X, Y), cons_2₂(X1, Y1))) -> if_samefringe_2_in_3_gg₇(U, V, X, Y, V1, Y1, samefringe_2_in_gg₂(V1, Y1))
if_samefringe_2_in_3_gg₇(U, V, X, Y, V1, Y1, samefringe_2_out_gg₂(V1, Y1)) -> samefringe_2_out_gg₂(cons_2₂(U, V), cons_2₂(X, Y))

The argument filtering Pi contains the following mapping:
samefringe_2_in_gg₂(x1, x2) = samefringe_2_in_gg₂(x1, x2)
nil_0 = nil_0
cons_2₂(x1, x2) = cons_2₂(x1, x2)
samefringe_2_out_gg₂(x1, x2) = samefringe_2_out_gg
if_samefringe_2_in_1_gg₅(x1, x2, x3, x4, x5) = if_samefringe_2_in_1_gg₃(x3, x4, x5)
gopher_2_in_ga₂(x1, x2) = gopher_2_in_ga₁(x1)
gopher_2_out_ga₂(x1, x2) = gopher_2_out_ga₁(x2)
if_gopher_2_in_1_ga₅(x1, x2, x3, x4, x5) = if_gopher_2_in_1_ga₁(x5)
if_samefringe_2_in_2_gg₆(x1, x2, x3, x4, x5, x6) = if_samefringe_2_in_2_gg₂(x5, x6)
if_samefringe_2_in_3_gg₇(x1, x2, x3, x4, x5, x6, x7) = if_samefringe_2_in_3_gg₁(x7)
IF_SAMEFRINGE_2_IN_2_GG₆(x1, x2, x3, x4, x5, x6) = IF_SAMEFRINGE_2_IN_2_GG₂(x5, x6)
IF_SAMEFRINGE_2_IN_3_GG₇(x1, x2, x3, x4, x5, x6, x7) = IF_SAMEFRINGE_2_IN_3_GG₁(x7)
SAMEFRINGE_2_IN_GG₂(x1, x2) = SAMEFRINGE_2_IN_GG₂(x1, x2)
IF_SAMEFRINGE_2_IN_1_GG₅(x1, x2, x3, x4, x5) = IF_SAMEFRINGE_2_IN_1_GG₃(x3, x4, x5)
GOPHER_2_IN_GA₂(x1, x2) = GOPHER_2_IN_GA₁(x1)
IF_GOPHER_2_IN_1_GA₅(x1, x2, x3, x4, x5) = IF_GOPHER_2_IN_1_GA₁(x5)

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:

SAMEFRINGE_2_IN_GG₂(cons_2₂(U, V), cons_2₂(X, Y)) -> IF_SAMEFRINGE_2_IN_1_GG₅(U, V, X, Y, gopher_2_in_ga₂(cons_2₂(U, V), cons_2₂(U1, V1)))
SAMEFRINGE_2_IN_GG₂(cons_2₂(U, V), cons_2₂(X, Y)) -> GOPHER_2_IN_GA₂(cons_2₂(U, V), cons_2₂(U1, V1))
GOPHER_2_IN_GA₂(cons_2₂(cons_2₂(U, V), W), X) -> IF_GOPHER_2_IN_1_GA₅(U, V, W, X, gopher_2_in_ga₂(cons_2₂(U, cons_2₂(V, W)), X))
GOPHER_2_IN_GA₂(cons_2₂(cons_2₂(U, V), W), X) -> GOPHER_2_IN_GA₂(cons_2₂(U, cons_2₂(V, W)), X)
IF_SAMEFRINGE_2_IN_1_GG₅(U, V, X, Y, gopher_2_out_ga₂(cons_2₂(U, V), cons_2₂(U1, V1))) -> IF_SAMEFRINGE_2_IN_2_GG₆(U, V, X, Y, V1, gopher_2_in_ga₂(cons_2₂(X, Y), cons_2₂(X1, Y1)))
IF_SAMEFRINGE_2_IN_1_GG₅(U, V, X, Y, gopher_2_out_ga₂(cons_2₂(U, V), cons_2₂(U1, V1))) -> GOPHER_2_IN_GA₂(cons_2₂(X, Y), cons_2₂(X1, Y1))
IF_SAMEFRINGE_2_IN_2_GG₆(U, V, X, Y, V1, gopher_2_out_ga₂(cons_2₂(X, Y), cons_2₂(X1, Y1))) -> IF_SAMEFRINGE_2_IN_3_GG₇(U, V, X, Y, V1, Y1, samefringe_2_in_gg₂(V1, Y1))
IF_SAMEFRINGE_2_IN_2_GG₆(U, V, X, Y, V1, gopher_2_out_ga₂(cons_2₂(X, Y), cons_2₂(X1, Y1))) -> SAMEFRINGE_2_IN_GG₂(V1, Y1)

The TRS R consists of the following rules:

samefringe_2_in_gg₂(nil_0, nil_0) -> samefringe_2_out_gg₂(nil_0, nil_0)
samefringe_2_in_gg₂(cons_2₂(U, V), cons_2₂(X, Y)) -> if_samefringe_2_in_1_gg₅(U, V, X, Y, gopher_2_in_ga₂(cons_2₂(U, V), cons_2₂(U1, V1)))
gopher_2_in_ga₂(nil_0, nil_0) -> gopher_2_out_ga₂(nil_0, nil_0)
gopher_2_in_ga₂(cons_2₂(nil_0, Y), cons_2₂(nil_0, Y)) -> gopher_2_out_ga₂(cons_2₂(nil_0, Y), cons_2₂(nil_0, Y))
gopher_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), X) -> if_gopher_2_in_1_ga₅(U, V, W, X, gopher_2_in_ga₂(cons_2₂(U, cons_2₂(V, W)), X))
if_gopher_2_in_1_ga₅(U, V, W, X, gopher_2_out_ga₂(cons_2₂(U, cons_2₂(V, W)), X)) -> gopher_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)
if_samefringe_2_in_1_gg₅(U, V, X, Y, gopher_2_out_ga₂(cons_2₂(U, V), cons_2₂(U1, V1))) -> if_samefringe_2_in_2_gg₆(U, V, X, Y, V1, gopher_2_in_ga₂(cons_2₂(X, Y), cons_2₂(X1, Y1)))
if_samefringe_2_in_2_gg₆(U, V, X, Y, V1, gopher_2_out_ga₂(cons_2₂(X, Y), cons_2₂(X1, Y1))) -> if_samefringe_2_in_3_gg₇(U, V, X, Y, V1, Y1, samefringe_2_in_gg₂(V1, Y1))
if_samefringe_2_in_3_gg₇(U, V, X, Y, V1, Y1, samefringe_2_out_gg₂(V1, Y1)) -> samefringe_2_out_gg₂(cons_2₂(U, V), cons_2₂(X, Y))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

GOPHER_2_IN_GA₂(cons_2₂(cons_2₂(U, V), W), X) -> GOPHER_2_IN_GA₂(cons_2₂(U, cons_2₂(V, W)), X)

The TRS R consists of the following rules:

samefringe_2_in_gg₂(nil_0, nil_0) -> samefringe_2_out_gg₂(nil_0, nil_0)
samefringe_2_in_gg₂(cons_2₂(U, V), cons_2₂(X, Y)) -> if_samefringe_2_in_1_gg₅(U, V, X, Y, gopher_2_in_ga₂(cons_2₂(U, V), cons_2₂(U1, V1)))
gopher_2_in_ga₂(nil_0, nil_0) -> gopher_2_out_ga₂(nil_0, nil_0)
gopher_2_in_ga₂(cons_2₂(nil_0, Y), cons_2₂(nil_0, Y)) -> gopher_2_out_ga₂(cons_2₂(nil_0, Y), cons_2₂(nil_0, Y))
gopher_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), X) -> if_gopher_2_in_1_ga₅(U, V, W, X, gopher_2_in_ga₂(cons_2₂(U, cons_2₂(V, W)), X))
if_gopher_2_in_1_ga₅(U, V, W, X, gopher_2_out_ga₂(cons_2₂(U, cons_2₂(V, W)), X)) -> gopher_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)
if_samefringe_2_in_1_gg₅(U, V, X, Y, gopher_2_out_ga₂(cons_2₂(U, V), cons_2₂(U1, V1))) -> if_samefringe_2_in_2_gg₆(U, V, X, Y, V1, gopher_2_in_ga₂(cons_2₂(X, Y), cons_2₂(X1, Y1)))
if_samefringe_2_in_2_gg₆(U, V, X, Y, V1, gopher_2_out_ga₂(cons_2₂(X, Y), cons_2₂(X1, Y1))) -> if_samefringe_2_in_3_gg₇(U, V, X, Y, V1, Y1, samefringe_2_in_gg₂(V1, Y1))
if_samefringe_2_in_3_gg₇(U, V, X, Y, V1, Y1, samefringe_2_out_gg₂(V1, Y1)) -> samefringe_2_out_gg₂(cons_2₂(U, V), cons_2₂(X, Y))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

GOPHER_2_IN_GA₂(cons_2₂(cons_2₂(U, V), W), X) -> GOPHER_2_IN_GA₂(cons_2₂(U, cons_2₂(V, W)), X)

R is empty.
The argument filtering Pi contains the following mapping:
cons_2₂(x1, x2) = cons_2₂(x1, x2)
GOPHER_2_IN_GA₂(x1, x2) = GOPHER_2_IN_GA₁(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

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

GOPHER_2_IN_GA₁(cons_2₂(cons_2₂(U, V), W)) -> GOPHER_2_IN_GA₁(cons_2₂(U, cons_2₂(V, W)))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {GOPHER_2_IN_GA₁}.
We used the following order and afs together with the size-change analysis to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
cons_2₂(x1, x2) = cons_2₁(x1)

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

GOPHER_2_IN_GA₁(cons_2₂(cons_2₂(U, V), W)) -> GOPHER_2_IN_GA₁(cons_2₂(U, cons_2₂(V, W))) (allowed arguments on rhs = {1})
The graph contains the following edges 1 > 1

We oriented the following set of usable rules. none

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

SAMEFRINGE_2_IN_GG₂(cons_2₂(U, V), cons_2₂(X, Y)) -> IF_SAMEFRINGE_2_IN_1_GG₅(U, V, X, Y, gopher_2_in_ga₂(cons_2₂(U, V), cons_2₂(U1, V1)))
IF_SAMEFRINGE_2_IN_1_GG₅(U, V, X, Y, gopher_2_out_ga₂(cons_2₂(U, V), cons_2₂(U1, V1))) -> IF_SAMEFRINGE_2_IN_2_GG₆(U, V, X, Y, V1, gopher_2_in_ga₂(cons_2₂(X, Y), cons_2₂(X1, Y1)))
IF_SAMEFRINGE_2_IN_2_GG₆(U, V, X, Y, V1, gopher_2_out_ga₂(cons_2₂(X, Y), cons_2₂(X1, Y1))) -> SAMEFRINGE_2_IN_GG₂(V1, Y1)

The TRS R consists of the following rules:

samefringe_2_in_gg₂(nil_0, nil_0) -> samefringe_2_out_gg₂(nil_0, nil_0)
samefringe_2_in_gg₂(cons_2₂(U, V), cons_2₂(X, Y)) -> if_samefringe_2_in_1_gg₅(U, V, X, Y, gopher_2_in_ga₂(cons_2₂(U, V), cons_2₂(U1, V1)))
gopher_2_in_ga₂(nil_0, nil_0) -> gopher_2_out_ga₂(nil_0, nil_0)
gopher_2_in_ga₂(cons_2₂(nil_0, Y), cons_2₂(nil_0, Y)) -> gopher_2_out_ga₂(cons_2₂(nil_0, Y), cons_2₂(nil_0, Y))
gopher_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), X) -> if_gopher_2_in_1_ga₅(U, V, W, X, gopher_2_in_ga₂(cons_2₂(U, cons_2₂(V, W)), X))
if_gopher_2_in_1_ga₅(U, V, W, X, gopher_2_out_ga₂(cons_2₂(U, cons_2₂(V, W)), X)) -> gopher_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)
if_samefringe_2_in_1_gg₅(U, V, X, Y, gopher_2_out_ga₂(cons_2₂(U, V), cons_2₂(U1, V1))) -> if_samefringe_2_in_2_gg₆(U, V, X, Y, V1, gopher_2_in_ga₂(cons_2₂(X, Y), cons_2₂(X1, Y1)))
if_samefringe_2_in_2_gg₆(U, V, X, Y, V1, gopher_2_out_ga₂(cons_2₂(X, Y), cons_2₂(X1, Y1))) -> if_samefringe_2_in_3_gg₇(U, V, X, Y, V1, Y1, samefringe_2_in_gg₂(V1, Y1))
if_samefringe_2_in_3_gg₇(U, V, X, Y, V1, Y1, samefringe_2_out_gg₂(V1, Y1)) -> samefringe_2_out_gg₂(cons_2₂(U, V), cons_2₂(X, Y))

The argument filtering Pi contains the following mapping:
samefringe_2_in_gg₂(x1, x2) = samefringe_2_in_gg₂(x1, x2)
nil_0 = nil_0
cons_2₂(x1, x2) = cons_2₂(x1, x2)
samefringe_2_out_gg₂(x1, x2) = samefringe_2_out_gg
if_samefringe_2_in_1_gg₅(x1, x2, x3, x4, x5) = if_samefringe_2_in_1_gg₃(x3, x4, x5)
gopher_2_in_ga₂(x1, x2) = gopher_2_in_ga₁(x1)
gopher_2_out_ga₂(x1, x2) = gopher_2_out_ga₁(x2)
if_gopher_2_in_1_ga₅(x1, x2, x3, x4, x5) = if_gopher_2_in_1_ga₁(x5)
if_samefringe_2_in_2_gg₆(x1, x2, x3, x4, x5, x6) = if_samefringe_2_in_2_gg₂(x5, x6)
if_samefringe_2_in_3_gg₇(x1, x2, x3, x4, x5, x6, x7) = if_samefringe_2_in_3_gg₁(x7)
IF_SAMEFRINGE_2_IN_2_GG₆(x1, x2, x3, x4, x5, x6) = IF_SAMEFRINGE_2_IN_2_GG₂(x5, x6)
SAMEFRINGE_2_IN_GG₂(x1, x2) = SAMEFRINGE_2_IN_GG₂(x1, x2)
IF_SAMEFRINGE_2_IN_1_GG₅(x1, x2, x3, x4, x5) = IF_SAMEFRINGE_2_IN_1_GG₃(x3, x4, x5)

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

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

SAMEFRINGE_2_IN_GG₂(cons_2₂(U, V), cons_2₂(X, Y)) -> IF_SAMEFRINGE_2_IN_1_GG₅(U, V, X, Y, gopher_2_in_ga₂(cons_2₂(U, V), cons_2₂(U1, V1)))
IF_SAMEFRINGE_2_IN_1_GG₅(U, V, X, Y, gopher_2_out_ga₂(cons_2₂(U, V), cons_2₂(U1, V1))) -> IF_SAMEFRINGE_2_IN_2_GG₆(U, V, X, Y, V1, gopher_2_in_ga₂(cons_2₂(X, Y), cons_2₂(X1, Y1)))
IF_SAMEFRINGE_2_IN_2_GG₆(U, V, X, Y, V1, gopher_2_out_ga₂(cons_2₂(X, Y), cons_2₂(X1, Y1))) -> SAMEFRINGE_2_IN_GG₂(V1, Y1)

The TRS R consists of the following rules:

gopher_2_in_ga₂(cons_2₂(nil_0, Y), cons_2₂(nil_0, Y)) -> gopher_2_out_ga₂(cons_2₂(nil_0, Y), cons_2₂(nil_0, Y))
gopher_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), X) -> if_gopher_2_in_1_ga₅(U, V, W, X, gopher_2_in_ga₂(cons_2₂(U, cons_2₂(V, W)), X))
if_gopher_2_in_1_ga₅(U, V, W, X, gopher_2_out_ga₂(cons_2₂(U, cons_2₂(V, W)), X)) -> gopher_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)

The argument filtering Pi contains the following mapping:
nil_0 = nil_0
cons_2₂(x1, x2) = cons_2₂(x1, x2)
gopher_2_in_ga₂(x1, x2) = gopher_2_in_ga₁(x1)
gopher_2_out_ga₂(x1, x2) = gopher_2_out_ga₁(x2)
if_gopher_2_in_1_ga₅(x1, x2, x3, x4, x5) = if_gopher_2_in_1_ga₁(x5)
IF_SAMEFRINGE_2_IN_2_GG₆(x1, x2, x3, x4, x5, x6) = IF_SAMEFRINGE_2_IN_2_GG₂(x5, x6)
SAMEFRINGE_2_IN_GG₂(x1, x2) = SAMEFRINGE_2_IN_GG₂(x1, x2)
IF_SAMEFRINGE_2_IN_1_GG₅(x1, x2, x3, x4, x5) = IF_SAMEFRINGE_2_IN_1_GG₃(x3, x4, x5)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

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

SAMEFRINGE_2_IN_GG₂(cons_2₂(U, V), cons_2₂(X, Y)) -> IF_SAMEFRINGE_2_IN_1_GG₃(X, Y, gopher_2_in_ga₁(cons_2₂(U, V)))
IF_SAMEFRINGE_2_IN_1_GG₃(X, Y, gopher_2_out_ga₁(cons_2₂(U1, V1))) -> IF_SAMEFRINGE_2_IN_2_GG₂(V1, gopher_2_in_ga₁(cons_2₂(X, Y)))
IF_SAMEFRINGE_2_IN_2_GG₂(V1, gopher_2_out_ga₁(cons_2₂(X1, Y1))) -> SAMEFRINGE_2_IN_GG₂(V1, Y1)

The TRS R consists of the following rules:

gopher_2_in_ga₁(cons_2₂(nil_0, Y)) -> gopher_2_out_ga₁(cons_2₂(nil_0, Y))
gopher_2_in_ga₁(cons_2₂(cons_2₂(U, V), W)) -> if_gopher_2_in_1_ga₁(gopher_2_in_ga₁(cons_2₂(U, cons_2₂(V, W))))
if_gopher_2_in_1_ga₁(gopher_2_out_ga₁(X)) -> gopher_2_out_ga₁(X)

The set Q consists of the following terms:

gopher_2_in_ga₁(x0)
if_gopher_2_in_1_ga₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_SAMEFRINGE_2_IN_1_GG₃, SAMEFRINGE_2_IN_GG₂, IF_SAMEFRINGE_2_IN_2_GG₂}.
We used the following order together with the size-change analysis to show that there are no infinite chains for this DP problem.

Order:Polynomial interpretation:

POL(nil_0) = 0
POL(if_gopher_2_in_1_ga₁(x₁)) = x₁
POL(gopher_2_out_ga₁(x₁)) = x₁
POL(gopher_2_in_ga₁(x₁)) = x₁
POL(cons_2₂(x₁, x₂)) = 1 + x₁ + x₂

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

IF_SAMEFRINGE_2_IN_1_GG₃(X, Y, gopher_2_out_ga₁(cons_2₂(U1, V1))) -> IF_SAMEFRINGE_2_IN_2_GG₂(V1, gopher_2_in_ga₁(cons_2₂(X, Y))) (allowed arguments on rhs = {1})
The graph contains the following edges 3 > 1
IF_SAMEFRINGE_2_IN_2_GG₂(V1, gopher_2_out_ga₁(cons_2₂(X1, Y1))) -> SAMEFRINGE_2_IN_GG₂(V1, Y1) (allowed arguments on rhs = {1, 2})
The graph contains the following edges 1 >= 1, 2 > 2
SAMEFRINGE_2_IN_GG₂(cons_2₂(U, V), cons_2₂(X, Y)) -> IF_SAMEFRINGE_2_IN_1_GG₃(X, Y, gopher_2_in_ga₁(cons_2₂(U, V))) (allowed arguments on rhs = {1, 2, 3})
The graph contains the following edges 2 > 1, 2 > 2, 1 >= 3

We oriented the following set of usable rules.

if_gopher_2_in_1_ga₁(gopher_2_out_ga₁(X)) -> gopher_2_out_ga₁(X)
gopher_2_in_ga₁(cons_2₂(nil_0, Y)) -> gopher_2_out_ga₁(cons_2₂(nil_0, Y))
gopher_2_in_ga₁(cons_2₂(cons_2₂(U, V), W)) -> if_gopher_2_in_1_ga₁(gopher_2_in_ga₁(cons_2₂(U, cons_2₂(V, W))))