Left Termination proof of /tmp/tmpU62eBx/perm.pl

Left Termination of the query pattern perm1(b,b) w.r.t. the given Prolog program could successfully be proven:

↳ PROLOG

  ↳ PrologToPiTRSProof

perm1₂(L, M) :- eq_len1₂(L, M), same_sets₂(L, M).
eq_len1₂({}₀, {}₀).
eq_len1₂(.₂(underscore, Xs), .₂(underscore1, Ys)) :- eq_len1₂(Xs, Ys).
member₂(X, .₂(X, underscore2)).
member₂(X, .₂(underscore3, T)) :- member₂(X, T).
same_sets₂({}₀, underscore4).
same_sets₂(.₂(X, Xs), L) :- member₂(X, L), same_sets₂(Xs, L).

With regard to the inferred argument filtering the predicates were used in the following modes:
perm1₂: (b,b)
eq_len1₂: (b,b)
same_sets₂: (b,b)
member₂: (b,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

perm1_2_in_gg₂(L, M) -> if_perm1_2_in_1_gg₃(L, M, eq_len1_2_in_gg₂(L, M))
eq_len1_2_in_gg₂([]_0, []_0) -> eq_len1_2_out_gg₂([]_0, []_0)
eq_len1_2_in_gg₂(._2₂(underscore, Xs), ._2₂(underscore1, Ys)) -> if_eq_len1_2_in_1_gg₅(underscore, Xs, underscore1, Ys, eq_len1_2_in_gg₂(Xs, Ys))
if_eq_len1_2_in_1_gg₅(underscore, Xs, underscore1, Ys, eq_len1_2_out_gg₂(Xs, Ys)) -> eq_len1_2_out_gg₂(._2₂(underscore, Xs), ._2₂(underscore1, Ys))
if_perm1_2_in_1_gg₃(L, M, eq_len1_2_out_gg₂(L, M)) -> if_perm1_2_in_2_gg₃(L, M, same_sets_2_in_gg₂(L, M))
same_sets_2_in_gg₂([]_0, underscore4) -> same_sets_2_out_gg₂([]_0, underscore4)
same_sets_2_in_gg₂(._2₂(X, Xs), L) -> if_same_sets_2_in_1_gg₄(X, Xs, L, member_2_in_gg₂(X, L))
member_2_in_gg₂(X, ._2₂(X, underscore2)) -> member_2_out_gg₂(X, ._2₂(X, underscore2))
member_2_in_gg₂(X, ._2₂(underscore3, T)) -> if_member_2_in_1_gg₄(X, underscore3, T, member_2_in_gg₂(X, T))
if_member_2_in_1_gg₄(X, underscore3, T, member_2_out_gg₂(X, T)) -> member_2_out_gg₂(X, ._2₂(underscore3, T))
if_same_sets_2_in_1_gg₄(X, Xs, L, member_2_out_gg₂(X, L)) -> if_same_sets_2_in_2_gg₄(X, Xs, L, same_sets_2_in_gg₂(Xs, L))
if_same_sets_2_in_2_gg₄(X, Xs, L, same_sets_2_out_gg₂(Xs, L)) -> same_sets_2_out_gg₂(._2₂(X, Xs), L)
if_perm1_2_in_2_gg₃(L, M, same_sets_2_out_gg₂(L, M)) -> perm1_2_out_gg₂(L, M)

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:

perm1_2_in_gg₂(L, M) -> if_perm1_2_in_1_gg₃(L, M, eq_len1_2_in_gg₂(L, M))
eq_len1_2_in_gg₂([]_0, []_0) -> eq_len1_2_out_gg₂([]_0, []_0)
eq_len1_2_in_gg₂(._2₂(underscore, Xs), ._2₂(underscore1, Ys)) -> if_eq_len1_2_in_1_gg₅(underscore, Xs, underscore1, Ys, eq_len1_2_in_gg₂(Xs, Ys))
if_eq_len1_2_in_1_gg₅(underscore, Xs, underscore1, Ys, eq_len1_2_out_gg₂(Xs, Ys)) -> eq_len1_2_out_gg₂(._2₂(underscore, Xs), ._2₂(underscore1, Ys))
if_perm1_2_in_1_gg₃(L, M, eq_len1_2_out_gg₂(L, M)) -> if_perm1_2_in_2_gg₃(L, M, same_sets_2_in_gg₂(L, M))
same_sets_2_in_gg₂([]_0, underscore4) -> same_sets_2_out_gg₂([]_0, underscore4)
same_sets_2_in_gg₂(._2₂(X, Xs), L) -> if_same_sets_2_in_1_gg₄(X, Xs, L, member_2_in_gg₂(X, L))
member_2_in_gg₂(X, ._2₂(X, underscore2)) -> member_2_out_gg₂(X, ._2₂(X, underscore2))
member_2_in_gg₂(X, ._2₂(underscore3, T)) -> if_member_2_in_1_gg₄(X, underscore3, T, member_2_in_gg₂(X, T))
if_member_2_in_1_gg₄(X, underscore3, T, member_2_out_gg₂(X, T)) -> member_2_out_gg₂(X, ._2₂(underscore3, T))
if_same_sets_2_in_1_gg₄(X, Xs, L, member_2_out_gg₂(X, L)) -> if_same_sets_2_in_2_gg₄(X, Xs, L, same_sets_2_in_gg₂(Xs, L))
if_same_sets_2_in_2_gg₄(X, Xs, L, same_sets_2_out_gg₂(Xs, L)) -> same_sets_2_out_gg₂(._2₂(X, Xs), L)
if_perm1_2_in_2_gg₃(L, M, same_sets_2_out_gg₂(L, M)) -> perm1_2_out_gg₂(L, M)

PERM1_2_IN_GG₂(L, M) -> IF_PERM1_2_IN_1_GG₃(L, M, eq_len1_2_in_gg₂(L, M))
PERM1_2_IN_GG₂(L, M) -> EQ_LEN1_2_IN_GG₂(L, M)
EQ_LEN1_2_IN_GG₂(._2₂(underscore, Xs), ._2₂(underscore1, Ys)) -> IF_EQ_LEN1_2_IN_1_GG₅(underscore, Xs, underscore1, Ys, eq_len1_2_in_gg₂(Xs, Ys))
EQ_LEN1_2_IN_GG₂(._2₂(underscore, Xs), ._2₂(underscore1, Ys)) -> EQ_LEN1_2_IN_GG₂(Xs, Ys)
IF_PERM1_2_IN_1_GG₃(L, M, eq_len1_2_out_gg₂(L, M)) -> IF_PERM1_2_IN_2_GG₃(L, M, same_sets_2_in_gg₂(L, M))
IF_PERM1_2_IN_1_GG₃(L, M, eq_len1_2_out_gg₂(L, M)) -> SAME_SETS_2_IN_GG₂(L, M)
SAME_SETS_2_IN_GG₂(._2₂(X, Xs), L) -> IF_SAME_SETS_2_IN_1_GG₄(X, Xs, L, member_2_in_gg₂(X, L))
SAME_SETS_2_IN_GG₂(._2₂(X, Xs), L) -> MEMBER_2_IN_GG₂(X, L)
MEMBER_2_IN_GG₂(X, ._2₂(underscore3, T)) -> IF_MEMBER_2_IN_1_GG₄(X, underscore3, T, member_2_in_gg₂(X, T))
MEMBER_2_IN_GG₂(X, ._2₂(underscore3, T)) -> MEMBER_2_IN_GG₂(X, T)
IF_SAME_SETS_2_IN_1_GG₄(X, Xs, L, member_2_out_gg₂(X, L)) -> IF_SAME_SETS_2_IN_2_GG₄(X, Xs, L, same_sets_2_in_gg₂(Xs, L))
IF_SAME_SETS_2_IN_1_GG₄(X, Xs, L, member_2_out_gg₂(X, L)) -> SAME_SETS_2_IN_GG₂(Xs, L)

The TRS R consists of the following rules:

perm1_2_in_gg₂(L, M) -> if_perm1_2_in_1_gg₃(L, M, eq_len1_2_in_gg₂(L, M))
eq_len1_2_in_gg₂([]_0, []_0) -> eq_len1_2_out_gg₂([]_0, []_0)
eq_len1_2_in_gg₂(._2₂(underscore, Xs), ._2₂(underscore1, Ys)) -> if_eq_len1_2_in_1_gg₅(underscore, Xs, underscore1, Ys, eq_len1_2_in_gg₂(Xs, Ys))
if_eq_len1_2_in_1_gg₅(underscore, Xs, underscore1, Ys, eq_len1_2_out_gg₂(Xs, Ys)) -> eq_len1_2_out_gg₂(._2₂(underscore, Xs), ._2₂(underscore1, Ys))
if_perm1_2_in_1_gg₃(L, M, eq_len1_2_out_gg₂(L, M)) -> if_perm1_2_in_2_gg₃(L, M, same_sets_2_in_gg₂(L, M))
same_sets_2_in_gg₂([]_0, underscore4) -> same_sets_2_out_gg₂([]_0, underscore4)
same_sets_2_in_gg₂(._2₂(X, Xs), L) -> if_same_sets_2_in_1_gg₄(X, Xs, L, member_2_in_gg₂(X, L))
member_2_in_gg₂(X, ._2₂(X, underscore2)) -> member_2_out_gg₂(X, ._2₂(X, underscore2))
member_2_in_gg₂(X, ._2₂(underscore3, T)) -> if_member_2_in_1_gg₄(X, underscore3, T, member_2_in_gg₂(X, T))
if_member_2_in_1_gg₄(X, underscore3, T, member_2_out_gg₂(X, T)) -> member_2_out_gg₂(X, ._2₂(underscore3, T))
if_same_sets_2_in_1_gg₄(X, Xs, L, member_2_out_gg₂(X, L)) -> if_same_sets_2_in_2_gg₄(X, Xs, L, same_sets_2_in_gg₂(Xs, L))
if_same_sets_2_in_2_gg₄(X, Xs, L, same_sets_2_out_gg₂(Xs, L)) -> same_sets_2_out_gg₂(._2₂(X, Xs), L)
if_perm1_2_in_2_gg₃(L, M, same_sets_2_out_gg₂(L, M)) -> perm1_2_out_gg₂(L, M)

The argument filtering Pi contains the following mapping:
perm1_2_in_gg₂(x1, x2) = perm1_2_in_gg₂(x1, x2)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
if_perm1_2_in_1_gg₃(x1, x2, x3) = if_perm1_2_in_1_gg₃(x1, x2, x3)
eq_len1_2_in_gg₂(x1, x2) = eq_len1_2_in_gg₂(x1, x2)
eq_len1_2_out_gg₂(x1, x2) = eq_len1_2_out_gg
if_eq_len1_2_in_1_gg₅(x1, x2, x3, x4, x5) = if_eq_len1_2_in_1_gg₁(x5)
if_perm1_2_in_2_gg₃(x1, x2, x3) = if_perm1_2_in_2_gg₁(x3)
same_sets_2_in_gg₂(x1, x2) = same_sets_2_in_gg₂(x1, x2)
same_sets_2_out_gg₂(x1, x2) = same_sets_2_out_gg
if_same_sets_2_in_1_gg₄(x1, x2, x3, x4) = if_same_sets_2_in_1_gg₃(x2, x3, x4)
member_2_in_gg₂(x1, x2) = member_2_in_gg₂(x1, x2)
member_2_out_gg₂(x1, x2) = member_2_out_gg
if_member_2_in_1_gg₄(x1, x2, x3, x4) = if_member_2_in_1_gg₁(x4)
if_same_sets_2_in_2_gg₄(x1, x2, x3, x4) = if_same_sets_2_in_2_gg₁(x4)
perm1_2_out_gg₂(x1, x2) = perm1_2_out_gg
IF_MEMBER_2_IN_1_GG₄(x1, x2, x3, x4) = IF_MEMBER_2_IN_1_GG₁(x4)
IF_EQ_LEN1_2_IN_1_GG₅(x1, x2, x3, x4, x5) = IF_EQ_LEN1_2_IN_1_GG₁(x5)
MEMBER_2_IN_GG₂(x1, x2) = MEMBER_2_IN_GG₂(x1, x2)
IF_PERM1_2_IN_1_GG₃(x1, x2, x3) = IF_PERM1_2_IN_1_GG₃(x1, x2, x3)
PERM1_2_IN_GG₂(x1, x2) = PERM1_2_IN_GG₂(x1, x2)
SAME_SETS_2_IN_GG₂(x1, x2) = SAME_SETS_2_IN_GG₂(x1, x2)
IF_SAME_SETS_2_IN_1_GG₄(x1, x2, x3, x4) = IF_SAME_SETS_2_IN_1_GG₃(x2, x3, x4)
IF_PERM1_2_IN_2_GG₃(x1, x2, x3) = IF_PERM1_2_IN_2_GG₁(x3)
IF_SAME_SETS_2_IN_2_GG₄(x1, x2, x3, x4) = IF_SAME_SETS_2_IN_2_GG₁(x4)
EQ_LEN1_2_IN_GG₂(x1, x2) = EQ_LEN1_2_IN_GG₂(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:

PERM1_2_IN_GG₂(L, M) -> IF_PERM1_2_IN_1_GG₃(L, M, eq_len1_2_in_gg₂(L, M))
PERM1_2_IN_GG₂(L, M) -> EQ_LEN1_2_IN_GG₂(L, M)
EQ_LEN1_2_IN_GG₂(._2₂(underscore, Xs), ._2₂(underscore1, Ys)) -> IF_EQ_LEN1_2_IN_1_GG₅(underscore, Xs, underscore1, Ys, eq_len1_2_in_gg₂(Xs, Ys))
EQ_LEN1_2_IN_GG₂(._2₂(underscore, Xs), ._2₂(underscore1, Ys)) -> EQ_LEN1_2_IN_GG₂(Xs, Ys)
IF_PERM1_2_IN_1_GG₃(L, M, eq_len1_2_out_gg₂(L, M)) -> IF_PERM1_2_IN_2_GG₃(L, M, same_sets_2_in_gg₂(L, M))
IF_PERM1_2_IN_1_GG₃(L, M, eq_len1_2_out_gg₂(L, M)) -> SAME_SETS_2_IN_GG₂(L, M)
SAME_SETS_2_IN_GG₂(._2₂(X, Xs), L) -> IF_SAME_SETS_2_IN_1_GG₄(X, Xs, L, member_2_in_gg₂(X, L))
SAME_SETS_2_IN_GG₂(._2₂(X, Xs), L) -> MEMBER_2_IN_GG₂(X, L)
MEMBER_2_IN_GG₂(X, ._2₂(underscore3, T)) -> IF_MEMBER_2_IN_1_GG₄(X, underscore3, T, member_2_in_gg₂(X, T))
MEMBER_2_IN_GG₂(X, ._2₂(underscore3, T)) -> MEMBER_2_IN_GG₂(X, T)
IF_SAME_SETS_2_IN_1_GG₄(X, Xs, L, member_2_out_gg₂(X, L)) -> IF_SAME_SETS_2_IN_2_GG₄(X, Xs, L, same_sets_2_in_gg₂(Xs, L))
IF_SAME_SETS_2_IN_1_GG₄(X, Xs, L, member_2_out_gg₂(X, L)) -> SAME_SETS_2_IN_GG₂(Xs, L)

The TRS R consists of the following rules:

perm1_2_in_gg₂(L, M) -> if_perm1_2_in_1_gg₃(L, M, eq_len1_2_in_gg₂(L, M))
eq_len1_2_in_gg₂([]_0, []_0) -> eq_len1_2_out_gg₂([]_0, []_0)
eq_len1_2_in_gg₂(._2₂(underscore, Xs), ._2₂(underscore1, Ys)) -> if_eq_len1_2_in_1_gg₅(underscore, Xs, underscore1, Ys, eq_len1_2_in_gg₂(Xs, Ys))
if_eq_len1_2_in_1_gg₅(underscore, Xs, underscore1, Ys, eq_len1_2_out_gg₂(Xs, Ys)) -> eq_len1_2_out_gg₂(._2₂(underscore, Xs), ._2₂(underscore1, Ys))
if_perm1_2_in_1_gg₃(L, M, eq_len1_2_out_gg₂(L, M)) -> if_perm1_2_in_2_gg₃(L, M, same_sets_2_in_gg₂(L, M))
same_sets_2_in_gg₂([]_0, underscore4) -> same_sets_2_out_gg₂([]_0, underscore4)
same_sets_2_in_gg₂(._2₂(X, Xs), L) -> if_same_sets_2_in_1_gg₄(X, Xs, L, member_2_in_gg₂(X, L))
member_2_in_gg₂(X, ._2₂(X, underscore2)) -> member_2_out_gg₂(X, ._2₂(X, underscore2))
member_2_in_gg₂(X, ._2₂(underscore3, T)) -> if_member_2_in_1_gg₄(X, underscore3, T, member_2_in_gg₂(X, T))
if_member_2_in_1_gg₄(X, underscore3, T, member_2_out_gg₂(X, T)) -> member_2_out_gg₂(X, ._2₂(underscore3, T))
if_same_sets_2_in_1_gg₄(X, Xs, L, member_2_out_gg₂(X, L)) -> if_same_sets_2_in_2_gg₄(X, Xs, L, same_sets_2_in_gg₂(Xs, L))
if_same_sets_2_in_2_gg₄(X, Xs, L, same_sets_2_out_gg₂(Xs, L)) -> same_sets_2_out_gg₂(._2₂(X, Xs), L)
if_perm1_2_in_2_gg₃(L, M, same_sets_2_out_gg₂(L, M)) -> perm1_2_out_gg₂(L, M)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

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

MEMBER_2_IN_GG₂(X, ._2₂(underscore3, T)) -> MEMBER_2_IN_GG₂(X, T)

The TRS R consists of the following rules:

perm1_2_in_gg₂(L, M) -> if_perm1_2_in_1_gg₃(L, M, eq_len1_2_in_gg₂(L, M))
eq_len1_2_in_gg₂([]_0, []_0) -> eq_len1_2_out_gg₂([]_0, []_0)
eq_len1_2_in_gg₂(._2₂(underscore, Xs), ._2₂(underscore1, Ys)) -> if_eq_len1_2_in_1_gg₅(underscore, Xs, underscore1, Ys, eq_len1_2_in_gg₂(Xs, Ys))
if_eq_len1_2_in_1_gg₅(underscore, Xs, underscore1, Ys, eq_len1_2_out_gg₂(Xs, Ys)) -> eq_len1_2_out_gg₂(._2₂(underscore, Xs), ._2₂(underscore1, Ys))
if_perm1_2_in_1_gg₃(L, M, eq_len1_2_out_gg₂(L, M)) -> if_perm1_2_in_2_gg₃(L, M, same_sets_2_in_gg₂(L, M))
same_sets_2_in_gg₂([]_0, underscore4) -> same_sets_2_out_gg₂([]_0, underscore4)
same_sets_2_in_gg₂(._2₂(X, Xs), L) -> if_same_sets_2_in_1_gg₄(X, Xs, L, member_2_in_gg₂(X, L))
member_2_in_gg₂(X, ._2₂(X, underscore2)) -> member_2_out_gg₂(X, ._2₂(X, underscore2))
member_2_in_gg₂(X, ._2₂(underscore3, T)) -> if_member_2_in_1_gg₄(X, underscore3, T, member_2_in_gg₂(X, T))
if_member_2_in_1_gg₄(X, underscore3, T, member_2_out_gg₂(X, T)) -> member_2_out_gg₂(X, ._2₂(underscore3, T))
if_same_sets_2_in_1_gg₄(X, Xs, L, member_2_out_gg₂(X, L)) -> if_same_sets_2_in_2_gg₄(X, Xs, L, same_sets_2_in_gg₂(Xs, L))
if_same_sets_2_in_2_gg₄(X, Xs, L, same_sets_2_out_gg₂(Xs, L)) -> same_sets_2_out_gg₂(._2₂(X, Xs), L)
if_perm1_2_in_2_gg₃(L, M, same_sets_2_out_gg₂(L, M)) -> perm1_2_out_gg₂(L, M)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

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

MEMBER_2_IN_GG₂(X, ._2₂(underscore3, T)) -> MEMBER_2_IN_GG₂(X, T)

R is empty.
Pi is empty.
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:

MEMBER_2_IN_GG₂(X, ._2₂(underscore3, T)) -> MEMBER_2_IN_GG₂(X, T)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {MEMBER_2_IN_GG₂}.
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:

MEMBER_2_IN_GG₂(X, ._2₂(underscore3, T)) -> MEMBER_2_IN_GG₂(X, T)
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:

IF_SAME_SETS_2_IN_1_GG₄(X, Xs, L, member_2_out_gg₂(X, L)) -> SAME_SETS_2_IN_GG₂(Xs, L)
SAME_SETS_2_IN_GG₂(._2₂(X, Xs), L) -> IF_SAME_SETS_2_IN_1_GG₄(X, Xs, L, member_2_in_gg₂(X, L))

The TRS R consists of the following rules:

perm1_2_in_gg₂(L, M) -> if_perm1_2_in_1_gg₃(L, M, eq_len1_2_in_gg₂(L, M))
eq_len1_2_in_gg₂([]_0, []_0) -> eq_len1_2_out_gg₂([]_0, []_0)
eq_len1_2_in_gg₂(._2₂(underscore, Xs), ._2₂(underscore1, Ys)) -> if_eq_len1_2_in_1_gg₅(underscore, Xs, underscore1, Ys, eq_len1_2_in_gg₂(Xs, Ys))
if_eq_len1_2_in_1_gg₅(underscore, Xs, underscore1, Ys, eq_len1_2_out_gg₂(Xs, Ys)) -> eq_len1_2_out_gg₂(._2₂(underscore, Xs), ._2₂(underscore1, Ys))
if_perm1_2_in_1_gg₃(L, M, eq_len1_2_out_gg₂(L, M)) -> if_perm1_2_in_2_gg₃(L, M, same_sets_2_in_gg₂(L, M))
same_sets_2_in_gg₂([]_0, underscore4) -> same_sets_2_out_gg₂([]_0, underscore4)
same_sets_2_in_gg₂(._2₂(X, Xs), L) -> if_same_sets_2_in_1_gg₄(X, Xs, L, member_2_in_gg₂(X, L))
member_2_in_gg₂(X, ._2₂(X, underscore2)) -> member_2_out_gg₂(X, ._2₂(X, underscore2))
member_2_in_gg₂(X, ._2₂(underscore3, T)) -> if_member_2_in_1_gg₄(X, underscore3, T, member_2_in_gg₂(X, T))
if_member_2_in_1_gg₄(X, underscore3, T, member_2_out_gg₂(X, T)) -> member_2_out_gg₂(X, ._2₂(underscore3, T))
if_same_sets_2_in_1_gg₄(X, Xs, L, member_2_out_gg₂(X, L)) -> if_same_sets_2_in_2_gg₄(X, Xs, L, same_sets_2_in_gg₂(Xs, L))
if_same_sets_2_in_2_gg₄(X, Xs, L, same_sets_2_out_gg₂(Xs, L)) -> same_sets_2_out_gg₂(._2₂(X, Xs), L)
if_perm1_2_in_2_gg₃(L, M, same_sets_2_out_gg₂(L, M)) -> perm1_2_out_gg₂(L, M)

The argument filtering Pi contains the following mapping:
perm1_2_in_gg₂(x1, x2) = perm1_2_in_gg₂(x1, x2)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
if_perm1_2_in_1_gg₃(x1, x2, x3) = if_perm1_2_in_1_gg₃(x1, x2, x3)
eq_len1_2_in_gg₂(x1, x2) = eq_len1_2_in_gg₂(x1, x2)
eq_len1_2_out_gg₂(x1, x2) = eq_len1_2_out_gg
if_eq_len1_2_in_1_gg₅(x1, x2, x3, x4, x5) = if_eq_len1_2_in_1_gg₁(x5)
if_perm1_2_in_2_gg₃(x1, x2, x3) = if_perm1_2_in_2_gg₁(x3)
same_sets_2_in_gg₂(x1, x2) = same_sets_2_in_gg₂(x1, x2)
same_sets_2_out_gg₂(x1, x2) = same_sets_2_out_gg
if_same_sets_2_in_1_gg₄(x1, x2, x3, x4) = if_same_sets_2_in_1_gg₃(x2, x3, x4)
member_2_in_gg₂(x1, x2) = member_2_in_gg₂(x1, x2)
member_2_out_gg₂(x1, x2) = member_2_out_gg
if_member_2_in_1_gg₄(x1, x2, x3, x4) = if_member_2_in_1_gg₁(x4)
if_same_sets_2_in_2_gg₄(x1, x2, x3, x4) = if_same_sets_2_in_2_gg₁(x4)
perm1_2_out_gg₂(x1, x2) = perm1_2_out_gg
SAME_SETS_2_IN_GG₂(x1, x2) = SAME_SETS_2_IN_GG₂(x1, x2)
IF_SAME_SETS_2_IN_1_GG₄(x1, x2, x3, x4) = IF_SAME_SETS_2_IN_1_GG₃(x2, 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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

IF_SAME_SETS_2_IN_1_GG₄(X, Xs, L, member_2_out_gg₂(X, L)) -> SAME_SETS_2_IN_GG₂(Xs, L)
SAME_SETS_2_IN_GG₂(._2₂(X, Xs), L) -> IF_SAME_SETS_2_IN_1_GG₄(X, Xs, L, member_2_in_gg₂(X, L))

The TRS R consists of the following rules:

member_2_in_gg₂(X, ._2₂(X, underscore2)) -> member_2_out_gg₂(X, ._2₂(X, underscore2))
member_2_in_gg₂(X, ._2₂(underscore3, T)) -> if_member_2_in_1_gg₄(X, underscore3, T, member_2_in_gg₂(X, T))
if_member_2_in_1_gg₄(X, underscore3, T, member_2_out_gg₂(X, T)) -> member_2_out_gg₂(X, ._2₂(underscore3, T))

The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
member_2_in_gg₂(x1, x2) = member_2_in_gg₂(x1, x2)
member_2_out_gg₂(x1, x2) = member_2_out_gg
if_member_2_in_1_gg₄(x1, x2, x3, x4) = if_member_2_in_1_gg₁(x4)
SAME_SETS_2_IN_GG₂(x1, x2) = SAME_SETS_2_IN_GG₂(x1, x2)
IF_SAME_SETS_2_IN_1_GG₄(x1, x2, x3, x4) = IF_SAME_SETS_2_IN_1_GG₃(x2, 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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

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

IF_SAME_SETS_2_IN_1_GG₃(Xs, L, member_2_out_gg) -> SAME_SETS_2_IN_GG₂(Xs, L)
SAME_SETS_2_IN_GG₂(._2₂(X, Xs), L) -> IF_SAME_SETS_2_IN_1_GG₃(Xs, L, member_2_in_gg₂(X, L))

The TRS R consists of the following rules:

member_2_in_gg₂(X, ._2₂(X, underscore2)) -> member_2_out_gg
member_2_in_gg₂(X, ._2₂(underscore3, T)) -> if_member_2_in_1_gg₁(member_2_in_gg₂(X, T))
if_member_2_in_1_gg₁(member_2_out_gg) -> member_2_out_gg

The set Q consists of the following terms:

member_2_in_gg₂(x0, x1)
if_member_2_in_1_gg₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {SAME_SETS_2_IN_GG₂, IF_SAME_SETS_2_IN_1_GG₃}.
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:

SAME_SETS_2_IN_GG₂(._2₂(X, Xs), L) -> IF_SAME_SETS_2_IN_1_GG₃(Xs, L, member_2_in_gg₂(X, L))
The graph contains the following edges 1 > 1, 2 >= 2
IF_SAME_SETS_2_IN_1_GG₃(Xs, L, member_2_out_gg) -> SAME_SETS_2_IN_GG₂(Xs, L)
The graph contains the following edges 1 >= 1, 2 >= 2

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

EQ_LEN1_2_IN_GG₂(._2₂(underscore, Xs), ._2₂(underscore1, Ys)) -> EQ_LEN1_2_IN_GG₂(Xs, Ys)

The TRS R consists of the following rules:

perm1_2_in_gg₂(L, M) -> if_perm1_2_in_1_gg₃(L, M, eq_len1_2_in_gg₂(L, M))
eq_len1_2_in_gg₂([]_0, []_0) -> eq_len1_2_out_gg₂([]_0, []_0)
eq_len1_2_in_gg₂(._2₂(underscore, Xs), ._2₂(underscore1, Ys)) -> if_eq_len1_2_in_1_gg₅(underscore, Xs, underscore1, Ys, eq_len1_2_in_gg₂(Xs, Ys))
if_eq_len1_2_in_1_gg₅(underscore, Xs, underscore1, Ys, eq_len1_2_out_gg₂(Xs, Ys)) -> eq_len1_2_out_gg₂(._2₂(underscore, Xs), ._2₂(underscore1, Ys))
if_perm1_2_in_1_gg₃(L, M, eq_len1_2_out_gg₂(L, M)) -> if_perm1_2_in_2_gg₃(L, M, same_sets_2_in_gg₂(L, M))
same_sets_2_in_gg₂([]_0, underscore4) -> same_sets_2_out_gg₂([]_0, underscore4)
same_sets_2_in_gg₂(._2₂(X, Xs), L) -> if_same_sets_2_in_1_gg₄(X, Xs, L, member_2_in_gg₂(X, L))
member_2_in_gg₂(X, ._2₂(X, underscore2)) -> member_2_out_gg₂(X, ._2₂(X, underscore2))
member_2_in_gg₂(X, ._2₂(underscore3, T)) -> if_member_2_in_1_gg₄(X, underscore3, T, member_2_in_gg₂(X, T))
if_member_2_in_1_gg₄(X, underscore3, T, member_2_out_gg₂(X, T)) -> member_2_out_gg₂(X, ._2₂(underscore3, T))
if_same_sets_2_in_1_gg₄(X, Xs, L, member_2_out_gg₂(X, L)) -> if_same_sets_2_in_2_gg₄(X, Xs, L, same_sets_2_in_gg₂(Xs, L))
if_same_sets_2_in_2_gg₄(X, Xs, L, same_sets_2_out_gg₂(Xs, L)) -> same_sets_2_out_gg₂(._2₂(X, Xs), L)
if_perm1_2_in_2_gg₃(L, M, same_sets_2_out_gg₂(L, M)) -> perm1_2_out_gg₂(L, M)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

EQ_LEN1_2_IN_GG₂(._2₂(underscore, Xs), ._2₂(underscore1, Ys)) -> EQ_LEN1_2_IN_GG₂(Xs, Ys)

R is empty.
Pi is empty.
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

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

EQ_LEN1_2_IN_GG₂(._2₂(underscore, Xs), ._2₂(underscore1, Ys)) -> EQ_LEN1_2_IN_GG₂(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 {EQ_LEN1_2_IN_GG₂}.
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:

EQ_LEN1_2_IN_GG₂(._2₂(underscore, Xs), ._2₂(underscore1, Ys)) -> EQ_LEN1_2_IN_GG₂(Xs, Ys)
The graph contains the following edges 1 > 1, 2 > 2