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

(0) Obligation:

Clauses:

perm1(L, M) :- ','(eq_len1(L, M), same_sets(L, M)).
eq_len1([], []).
eq_len1(.(X1, Xs), .(X2, Ys)) :- eq_len1(Xs, Ys).
member(X, .(X, X3)).
member(X, .(X4, T)) :- member(X, T).
same_sets([], X5).
same_sets(.(X, Xs), L) :- ','(member(X, L), same_sets(Xs, L)).

Query: perm1(g,g)

(1) PrologToPiTRSProof (SOUND transformation)

We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
perm1_in: (b,b)
eq_len1_in: (b,b)
same_sets_in: (b,b)
member_in: (b,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

perm1_in_gg(L, M) → U1_gg(L, M, eq_len1_in_gg(L, M))
eq_len1_in_gg([], []) → eq_len1_out_gg([], [])
eq_len1_in_gg(.(X1, Xs), .(X2, Ys)) → U3_gg(X1, Xs, X2, Ys, eq_len1_in_gg(Xs, Ys))
U3_gg(X1, Xs, X2, Ys, eq_len1_out_gg(Xs, Ys)) → eq_len1_out_gg(.(X1, Xs), .(X2, Ys))
U1_gg(L, M, eq_len1_out_gg(L, M)) → U2_gg(L, M, same_sets_in_gg(L, M))
same_sets_in_gg([], X5) → same_sets_out_gg([], X5)
same_sets_in_gg(.(X, Xs), L) → U5_gg(X, Xs, L, member_in_gg(X, L))
member_in_gg(X, .(X, X3)) → member_out_gg(X, .(X, X3))
member_in_gg(X, .(X4, T)) → U4_gg(X, X4, T, member_in_gg(X, T))
U4_gg(X, X4, T, member_out_gg(X, T)) → member_out_gg(X, .(X4, T))
U5_gg(X, Xs, L, member_out_gg(X, L)) → U6_gg(X, Xs, L, same_sets_in_gg(Xs, L))
U6_gg(X, Xs, L, same_sets_out_gg(Xs, L)) → same_sets_out_gg(.(X, Xs), L)
U2_gg(L, M, same_sets_out_gg(L, M)) → perm1_out_gg(L, M)

The argument filtering Pi contains the following mapping:
perm1_in_gg(x1, x2)  =  perm1_in_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x2, x3)
eq_len1_in_gg(x1, x2)  =  eq_len1_in_gg(x1, x2)
[]  =  []
eq_len1_out_gg(x1, x2)  =  eq_len1_out_gg
.(x1, x2)  =  .(x1, x2)
U3_gg(x1, x2, x3, x4, x5)  =  U3_gg(x5)
U2_gg(x1, x2, x3)  =  U2_gg(x3)
same_sets_in_gg(x1, x2)  =  same_sets_in_gg(x1, x2)
same_sets_out_gg(x1, x2)  =  same_sets_out_gg
U5_gg(x1, x2, x3, x4)  =  U5_gg(x2, x3, x4)
member_in_gg(x1, x2)  =  member_in_gg(x1, x2)
member_out_gg(x1, x2)  =  member_out_gg
U4_gg(x1, x2, x3, x4)  =  U4_gg(x4)
U6_gg(x1, x2, x3, x4)  =  U6_gg(x4)
perm1_out_gg(x1, x2)  =  perm1_out_gg

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(2) Obligation:

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

perm1_in_gg(L, M) → U1_gg(L, M, eq_len1_in_gg(L, M))
eq_len1_in_gg([], []) → eq_len1_out_gg([], [])
eq_len1_in_gg(.(X1, Xs), .(X2, Ys)) → U3_gg(X1, Xs, X2, Ys, eq_len1_in_gg(Xs, Ys))
U3_gg(X1, Xs, X2, Ys, eq_len1_out_gg(Xs, Ys)) → eq_len1_out_gg(.(X1, Xs), .(X2, Ys))
U1_gg(L, M, eq_len1_out_gg(L, M)) → U2_gg(L, M, same_sets_in_gg(L, M))
same_sets_in_gg([], X5) → same_sets_out_gg([], X5)
same_sets_in_gg(.(X, Xs), L) → U5_gg(X, Xs, L, member_in_gg(X, L))
member_in_gg(X, .(X, X3)) → member_out_gg(X, .(X, X3))
member_in_gg(X, .(X4, T)) → U4_gg(X, X4, T, member_in_gg(X, T))
U4_gg(X, X4, T, member_out_gg(X, T)) → member_out_gg(X, .(X4, T))
U5_gg(X, Xs, L, member_out_gg(X, L)) → U6_gg(X, Xs, L, same_sets_in_gg(Xs, L))
U6_gg(X, Xs, L, same_sets_out_gg(Xs, L)) → same_sets_out_gg(.(X, Xs), L)
U2_gg(L, M, same_sets_out_gg(L, M)) → perm1_out_gg(L, M)

The argument filtering Pi contains the following mapping:
perm1_in_gg(x1, x2)  =  perm1_in_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x2, x3)
eq_len1_in_gg(x1, x2)  =  eq_len1_in_gg(x1, x2)
[]  =  []
eq_len1_out_gg(x1, x2)  =  eq_len1_out_gg
.(x1, x2)  =  .(x1, x2)
U3_gg(x1, x2, x3, x4, x5)  =  U3_gg(x5)
U2_gg(x1, x2, x3)  =  U2_gg(x3)
same_sets_in_gg(x1, x2)  =  same_sets_in_gg(x1, x2)
same_sets_out_gg(x1, x2)  =  same_sets_out_gg
U5_gg(x1, x2, x3, x4)  =  U5_gg(x2, x3, x4)
member_in_gg(x1, x2)  =  member_in_gg(x1, x2)
member_out_gg(x1, x2)  =  member_out_gg
U4_gg(x1, x2, x3, x4)  =  U4_gg(x4)
U6_gg(x1, x2, x3, x4)  =  U6_gg(x4)
perm1_out_gg(x1, x2)  =  perm1_out_gg

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

PERM1_IN_GG(L, M) → U1_GG(L, M, eq_len1_in_gg(L, M))
PERM1_IN_GG(L, M) → EQ_LEN1_IN_GG(L, M)
EQ_LEN1_IN_GG(.(X1, Xs), .(X2, Ys)) → U3_GG(X1, Xs, X2, Ys, eq_len1_in_gg(Xs, Ys))
EQ_LEN1_IN_GG(.(X1, Xs), .(X2, Ys)) → EQ_LEN1_IN_GG(Xs, Ys)
U1_GG(L, M, eq_len1_out_gg(L, M)) → U2_GG(L, M, same_sets_in_gg(L, M))
U1_GG(L, M, eq_len1_out_gg(L, M)) → SAME_SETS_IN_GG(L, M)
SAME_SETS_IN_GG(.(X, Xs), L) → U5_GG(X, Xs, L, member_in_gg(X, L))
SAME_SETS_IN_GG(.(X, Xs), L) → MEMBER_IN_GG(X, L)
MEMBER_IN_GG(X, .(X4, T)) → U4_GG(X, X4, T, member_in_gg(X, T))
MEMBER_IN_GG(X, .(X4, T)) → MEMBER_IN_GG(X, T)
U5_GG(X, Xs, L, member_out_gg(X, L)) → U6_GG(X, Xs, L, same_sets_in_gg(Xs, L))
U5_GG(X, Xs, L, member_out_gg(X, L)) → SAME_SETS_IN_GG(Xs, L)

The TRS R consists of the following rules:

perm1_in_gg(L, M) → U1_gg(L, M, eq_len1_in_gg(L, M))
eq_len1_in_gg([], []) → eq_len1_out_gg([], [])
eq_len1_in_gg(.(X1, Xs), .(X2, Ys)) → U3_gg(X1, Xs, X2, Ys, eq_len1_in_gg(Xs, Ys))
U3_gg(X1, Xs, X2, Ys, eq_len1_out_gg(Xs, Ys)) → eq_len1_out_gg(.(X1, Xs), .(X2, Ys))
U1_gg(L, M, eq_len1_out_gg(L, M)) → U2_gg(L, M, same_sets_in_gg(L, M))
same_sets_in_gg([], X5) → same_sets_out_gg([], X5)
same_sets_in_gg(.(X, Xs), L) → U5_gg(X, Xs, L, member_in_gg(X, L))
member_in_gg(X, .(X, X3)) → member_out_gg(X, .(X, X3))
member_in_gg(X, .(X4, T)) → U4_gg(X, X4, T, member_in_gg(X, T))
U4_gg(X, X4, T, member_out_gg(X, T)) → member_out_gg(X, .(X4, T))
U5_gg(X, Xs, L, member_out_gg(X, L)) → U6_gg(X, Xs, L, same_sets_in_gg(Xs, L))
U6_gg(X, Xs, L, same_sets_out_gg(Xs, L)) → same_sets_out_gg(.(X, Xs), L)
U2_gg(L, M, same_sets_out_gg(L, M)) → perm1_out_gg(L, M)

The argument filtering Pi contains the following mapping:
perm1_in_gg(x1, x2)  =  perm1_in_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x2, x3)
eq_len1_in_gg(x1, x2)  =  eq_len1_in_gg(x1, x2)
[]  =  []
eq_len1_out_gg(x1, x2)  =  eq_len1_out_gg
.(x1, x2)  =  .(x1, x2)
U3_gg(x1, x2, x3, x4, x5)  =  U3_gg(x5)
U2_gg(x1, x2, x3)  =  U2_gg(x3)
same_sets_in_gg(x1, x2)  =  same_sets_in_gg(x1, x2)
same_sets_out_gg(x1, x2)  =  same_sets_out_gg
U5_gg(x1, x2, x3, x4)  =  U5_gg(x2, x3, x4)
member_in_gg(x1, x2)  =  member_in_gg(x1, x2)
member_out_gg(x1, x2)  =  member_out_gg
U4_gg(x1, x2, x3, x4)  =  U4_gg(x4)
U6_gg(x1, x2, x3, x4)  =  U6_gg(x4)
perm1_out_gg(x1, x2)  =  perm1_out_gg
PERM1_IN_GG(x1, x2)  =  PERM1_IN_GG(x1, x2)
U1_GG(x1, x2, x3)  =  U1_GG(x1, x2, x3)
EQ_LEN1_IN_GG(x1, x2)  =  EQ_LEN1_IN_GG(x1, x2)
U3_GG(x1, x2, x3, x4, x5)  =  U3_GG(x5)
U2_GG(x1, x2, x3)  =  U2_GG(x3)
SAME_SETS_IN_GG(x1, x2)  =  SAME_SETS_IN_GG(x1, x2)
U5_GG(x1, x2, x3, x4)  =  U5_GG(x2, x3, x4)
MEMBER_IN_GG(x1, x2)  =  MEMBER_IN_GG(x1, x2)
U4_GG(x1, x2, x3, x4)  =  U4_GG(x4)
U6_GG(x1, x2, x3, x4)  =  U6_GG(x4)

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

(4) Obligation:

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

PERM1_IN_GG(L, M) → U1_GG(L, M, eq_len1_in_gg(L, M))
PERM1_IN_GG(L, M) → EQ_LEN1_IN_GG(L, M)
EQ_LEN1_IN_GG(.(X1, Xs), .(X2, Ys)) → U3_GG(X1, Xs, X2, Ys, eq_len1_in_gg(Xs, Ys))
EQ_LEN1_IN_GG(.(X1, Xs), .(X2, Ys)) → EQ_LEN1_IN_GG(Xs, Ys)
U1_GG(L, M, eq_len1_out_gg(L, M)) → U2_GG(L, M, same_sets_in_gg(L, M))
U1_GG(L, M, eq_len1_out_gg(L, M)) → SAME_SETS_IN_GG(L, M)
SAME_SETS_IN_GG(.(X, Xs), L) → U5_GG(X, Xs, L, member_in_gg(X, L))
SAME_SETS_IN_GG(.(X, Xs), L) → MEMBER_IN_GG(X, L)
MEMBER_IN_GG(X, .(X4, T)) → U4_GG(X, X4, T, member_in_gg(X, T))
MEMBER_IN_GG(X, .(X4, T)) → MEMBER_IN_GG(X, T)
U5_GG(X, Xs, L, member_out_gg(X, L)) → U6_GG(X, Xs, L, same_sets_in_gg(Xs, L))
U5_GG(X, Xs, L, member_out_gg(X, L)) → SAME_SETS_IN_GG(Xs, L)

The TRS R consists of the following rules:

perm1_in_gg(L, M) → U1_gg(L, M, eq_len1_in_gg(L, M))
eq_len1_in_gg([], []) → eq_len1_out_gg([], [])
eq_len1_in_gg(.(X1, Xs), .(X2, Ys)) → U3_gg(X1, Xs, X2, Ys, eq_len1_in_gg(Xs, Ys))
U3_gg(X1, Xs, X2, Ys, eq_len1_out_gg(Xs, Ys)) → eq_len1_out_gg(.(X1, Xs), .(X2, Ys))
U1_gg(L, M, eq_len1_out_gg(L, M)) → U2_gg(L, M, same_sets_in_gg(L, M))
same_sets_in_gg([], X5) → same_sets_out_gg([], X5)
same_sets_in_gg(.(X, Xs), L) → U5_gg(X, Xs, L, member_in_gg(X, L))
member_in_gg(X, .(X, X3)) → member_out_gg(X, .(X, X3))
member_in_gg(X, .(X4, T)) → U4_gg(X, X4, T, member_in_gg(X, T))
U4_gg(X, X4, T, member_out_gg(X, T)) → member_out_gg(X, .(X4, T))
U5_gg(X, Xs, L, member_out_gg(X, L)) → U6_gg(X, Xs, L, same_sets_in_gg(Xs, L))
U6_gg(X, Xs, L, same_sets_out_gg(Xs, L)) → same_sets_out_gg(.(X, Xs), L)
U2_gg(L, M, same_sets_out_gg(L, M)) → perm1_out_gg(L, M)

The argument filtering Pi contains the following mapping:
perm1_in_gg(x1, x2)  =  perm1_in_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x2, x3)
eq_len1_in_gg(x1, x2)  =  eq_len1_in_gg(x1, x2)
[]  =  []
eq_len1_out_gg(x1, x2)  =  eq_len1_out_gg
.(x1, x2)  =  .(x1, x2)
U3_gg(x1, x2, x3, x4, x5)  =  U3_gg(x5)
U2_gg(x1, x2, x3)  =  U2_gg(x3)
same_sets_in_gg(x1, x2)  =  same_sets_in_gg(x1, x2)
same_sets_out_gg(x1, x2)  =  same_sets_out_gg
U5_gg(x1, x2, x3, x4)  =  U5_gg(x2, x3, x4)
member_in_gg(x1, x2)  =  member_in_gg(x1, x2)
member_out_gg(x1, x2)  =  member_out_gg
U4_gg(x1, x2, x3, x4)  =  U4_gg(x4)
U6_gg(x1, x2, x3, x4)  =  U6_gg(x4)
perm1_out_gg(x1, x2)  =  perm1_out_gg
PERM1_IN_GG(x1, x2)  =  PERM1_IN_GG(x1, x2)
U1_GG(x1, x2, x3)  =  U1_GG(x1, x2, x3)
EQ_LEN1_IN_GG(x1, x2)  =  EQ_LEN1_IN_GG(x1, x2)
U3_GG(x1, x2, x3, x4, x5)  =  U3_GG(x5)
U2_GG(x1, x2, x3)  =  U2_GG(x3)
SAME_SETS_IN_GG(x1, x2)  =  SAME_SETS_IN_GG(x1, x2)
U5_GG(x1, x2, x3, x4)  =  U5_GG(x2, x3, x4)
MEMBER_IN_GG(x1, x2)  =  MEMBER_IN_GG(x1, x2)
U4_GG(x1, x2, x3, x4)  =  U4_GG(x4)
U6_GG(x1, x2, x3, x4)  =  U6_GG(x4)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 3 SCCs with 8 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

MEMBER_IN_GG(X, .(X4, T)) → MEMBER_IN_GG(X, T)

The TRS R consists of the following rules:

perm1_in_gg(L, M) → U1_gg(L, M, eq_len1_in_gg(L, M))
eq_len1_in_gg([], []) → eq_len1_out_gg([], [])
eq_len1_in_gg(.(X1, Xs), .(X2, Ys)) → U3_gg(X1, Xs, X2, Ys, eq_len1_in_gg(Xs, Ys))
U3_gg(X1, Xs, X2, Ys, eq_len1_out_gg(Xs, Ys)) → eq_len1_out_gg(.(X1, Xs), .(X2, Ys))
U1_gg(L, M, eq_len1_out_gg(L, M)) → U2_gg(L, M, same_sets_in_gg(L, M))
same_sets_in_gg([], X5) → same_sets_out_gg([], X5)
same_sets_in_gg(.(X, Xs), L) → U5_gg(X, Xs, L, member_in_gg(X, L))
member_in_gg(X, .(X, X3)) → member_out_gg(X, .(X, X3))
member_in_gg(X, .(X4, T)) → U4_gg(X, X4, T, member_in_gg(X, T))
U4_gg(X, X4, T, member_out_gg(X, T)) → member_out_gg(X, .(X4, T))
U5_gg(X, Xs, L, member_out_gg(X, L)) → U6_gg(X, Xs, L, same_sets_in_gg(Xs, L))
U6_gg(X, Xs, L, same_sets_out_gg(Xs, L)) → same_sets_out_gg(.(X, Xs), L)
U2_gg(L, M, same_sets_out_gg(L, M)) → perm1_out_gg(L, M)

The argument filtering Pi contains the following mapping:
perm1_in_gg(x1, x2)  =  perm1_in_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x2, x3)
eq_len1_in_gg(x1, x2)  =  eq_len1_in_gg(x1, x2)
[]  =  []
eq_len1_out_gg(x1, x2)  =  eq_len1_out_gg
.(x1, x2)  =  .(x1, x2)
U3_gg(x1, x2, x3, x4, x5)  =  U3_gg(x5)
U2_gg(x1, x2, x3)  =  U2_gg(x3)
same_sets_in_gg(x1, x2)  =  same_sets_in_gg(x1, x2)
same_sets_out_gg(x1, x2)  =  same_sets_out_gg
U5_gg(x1, x2, x3, x4)  =  U5_gg(x2, x3, x4)
member_in_gg(x1, x2)  =  member_in_gg(x1, x2)
member_out_gg(x1, x2)  =  member_out_gg
U4_gg(x1, x2, x3, x4)  =  U4_gg(x4)
U6_gg(x1, x2, x3, x4)  =  U6_gg(x4)
perm1_out_gg(x1, x2)  =  perm1_out_gg
MEMBER_IN_GG(x1, x2)  =  MEMBER_IN_GG(x1, x2)

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

(8) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(9) Obligation:

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

MEMBER_IN_GG(X, .(X4, T)) → MEMBER_IN_GG(X, T)

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

(10) PiDPToQDPProof (EQUIVALENT transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(11) Obligation:

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

MEMBER_IN_GG(X, .(X4, T)) → MEMBER_IN_GG(X, T)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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_IN_GG(X, .(X4, T)) → MEMBER_IN_GG(X, T)
  The graph contains the following edges 1 >= 1, 2 > 2

(13) YES

(14) Obligation:

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

U5_GG(X, Xs, L, member_out_gg(X, L)) → SAME_SETS_IN_GG(Xs, L)
SAME_SETS_IN_GG(.(X, Xs), L) → U5_GG(X, Xs, L, member_in_gg(X, L))

The TRS R consists of the following rules:

perm1_in_gg(L, M) → U1_gg(L, M, eq_len1_in_gg(L, M))
eq_len1_in_gg([], []) → eq_len1_out_gg([], [])
eq_len1_in_gg(.(X1, Xs), .(X2, Ys)) → U3_gg(X1, Xs, X2, Ys, eq_len1_in_gg(Xs, Ys))
U3_gg(X1, Xs, X2, Ys, eq_len1_out_gg(Xs, Ys)) → eq_len1_out_gg(.(X1, Xs), .(X2, Ys))
U1_gg(L, M, eq_len1_out_gg(L, M)) → U2_gg(L, M, same_sets_in_gg(L, M))
same_sets_in_gg([], X5) → same_sets_out_gg([], X5)
same_sets_in_gg(.(X, Xs), L) → U5_gg(X, Xs, L, member_in_gg(X, L))
member_in_gg(X, .(X, X3)) → member_out_gg(X, .(X, X3))
member_in_gg(X, .(X4, T)) → U4_gg(X, X4, T, member_in_gg(X, T))
U4_gg(X, X4, T, member_out_gg(X, T)) → member_out_gg(X, .(X4, T))
U5_gg(X, Xs, L, member_out_gg(X, L)) → U6_gg(X, Xs, L, same_sets_in_gg(Xs, L))
U6_gg(X, Xs, L, same_sets_out_gg(Xs, L)) → same_sets_out_gg(.(X, Xs), L)
U2_gg(L, M, same_sets_out_gg(L, M)) → perm1_out_gg(L, M)

The argument filtering Pi contains the following mapping:
perm1_in_gg(x1, x2)  =  perm1_in_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x2, x3)
eq_len1_in_gg(x1, x2)  =  eq_len1_in_gg(x1, x2)
[]  =  []
eq_len1_out_gg(x1, x2)  =  eq_len1_out_gg
.(x1, x2)  =  .(x1, x2)
U3_gg(x1, x2, x3, x4, x5)  =  U3_gg(x5)
U2_gg(x1, x2, x3)  =  U2_gg(x3)
same_sets_in_gg(x1, x2)  =  same_sets_in_gg(x1, x2)
same_sets_out_gg(x1, x2)  =  same_sets_out_gg
U5_gg(x1, x2, x3, x4)  =  U5_gg(x2, x3, x4)
member_in_gg(x1, x2)  =  member_in_gg(x1, x2)
member_out_gg(x1, x2)  =  member_out_gg
U4_gg(x1, x2, x3, x4)  =  U4_gg(x4)
U6_gg(x1, x2, x3, x4)  =  U6_gg(x4)
perm1_out_gg(x1, x2)  =  perm1_out_gg
SAME_SETS_IN_GG(x1, x2)  =  SAME_SETS_IN_GG(x1, x2)
U5_GG(x1, x2, x3, x4)  =  U5_GG(x2, x3, x4)

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

(15) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(16) Obligation:

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

U5_GG(X, Xs, L, member_out_gg(X, L)) → SAME_SETS_IN_GG(Xs, L)
SAME_SETS_IN_GG(.(X, Xs), L) → U5_GG(X, Xs, L, member_in_gg(X, L))

The TRS R consists of the following rules:

member_in_gg(X, .(X, X3)) → member_out_gg(X, .(X, X3))
member_in_gg(X, .(X4, T)) → U4_gg(X, X4, T, member_in_gg(X, T))
U4_gg(X, X4, T, member_out_gg(X, T)) → member_out_gg(X, .(X4, T))

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
member_in_gg(x1, x2)  =  member_in_gg(x1, x2)
member_out_gg(x1, x2)  =  member_out_gg
U4_gg(x1, x2, x3, x4)  =  U4_gg(x4)
SAME_SETS_IN_GG(x1, x2)  =  SAME_SETS_IN_GG(x1, x2)
U5_GG(x1, x2, x3, x4)  =  U5_GG(x2, x3, x4)

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

(17) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(18) Obligation:

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

U5_GG(Xs, L, member_out_gg) → SAME_SETS_IN_GG(Xs, L)
SAME_SETS_IN_GG(.(X, Xs), L) → U5_GG(Xs, L, member_in_gg(X, L))

The TRS R consists of the following rules:

member_in_gg(X, .(X, X3)) → member_out_gg
member_in_gg(X, .(X4, T)) → U4_gg(member_in_gg(X, T))
U4_gg(member_out_gg) → member_out_gg

The set Q consists of the following terms:

member_in_gg(x0, x1)
U4_gg(x0)

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

(19) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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_IN_GG(.(X, Xs), L) → U5_GG(Xs, L, member_in_gg(X, L))
  The graph contains the following edges 1 > 1, 2 >= 2

• U5_GG(Xs, L, member_out_gg) → SAME_SETS_IN_GG(Xs, L)
  The graph contains the following edges 1 >= 1, 2 >= 2

(20) YES

(21) Obligation:

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

EQ_LEN1_IN_GG(.(X1, Xs), .(X2, Ys)) → EQ_LEN1_IN_GG(Xs, Ys)

The TRS R consists of the following rules:

perm1_in_gg(L, M) → U1_gg(L, M, eq_len1_in_gg(L, M))
eq_len1_in_gg([], []) → eq_len1_out_gg([], [])
eq_len1_in_gg(.(X1, Xs), .(X2, Ys)) → U3_gg(X1, Xs, X2, Ys, eq_len1_in_gg(Xs, Ys))
U3_gg(X1, Xs, X2, Ys, eq_len1_out_gg(Xs, Ys)) → eq_len1_out_gg(.(X1, Xs), .(X2, Ys))
U1_gg(L, M, eq_len1_out_gg(L, M)) → U2_gg(L, M, same_sets_in_gg(L, M))
same_sets_in_gg([], X5) → same_sets_out_gg([], X5)
same_sets_in_gg(.(X, Xs), L) → U5_gg(X, Xs, L, member_in_gg(X, L))
member_in_gg(X, .(X, X3)) → member_out_gg(X, .(X, X3))
member_in_gg(X, .(X4, T)) → U4_gg(X, X4, T, member_in_gg(X, T))
U4_gg(X, X4, T, member_out_gg(X, T)) → member_out_gg(X, .(X4, T))
U5_gg(X, Xs, L, member_out_gg(X, L)) → U6_gg(X, Xs, L, same_sets_in_gg(Xs, L))
U6_gg(X, Xs, L, same_sets_out_gg(Xs, L)) → same_sets_out_gg(.(X, Xs), L)
U2_gg(L, M, same_sets_out_gg(L, M)) → perm1_out_gg(L, M)

The argument filtering Pi contains the following mapping:
perm1_in_gg(x1, x2)  =  perm1_in_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x2, x3)
eq_len1_in_gg(x1, x2)  =  eq_len1_in_gg(x1, x2)
[]  =  []
eq_len1_out_gg(x1, x2)  =  eq_len1_out_gg
.(x1, x2)  =  .(x1, x2)
U3_gg(x1, x2, x3, x4, x5)  =  U3_gg(x5)
U2_gg(x1, x2, x3)  =  U2_gg(x3)
same_sets_in_gg(x1, x2)  =  same_sets_in_gg(x1, x2)
same_sets_out_gg(x1, x2)  =  same_sets_out_gg
U5_gg(x1, x2, x3, x4)  =  U5_gg(x2, x3, x4)
member_in_gg(x1, x2)  =  member_in_gg(x1, x2)
member_out_gg(x1, x2)  =  member_out_gg
U4_gg(x1, x2, x3, x4)  =  U4_gg(x4)
U6_gg(x1, x2, x3, x4)  =  U6_gg(x4)
perm1_out_gg(x1, x2)  =  perm1_out_gg
EQ_LEN1_IN_GG(x1, x2)  =  EQ_LEN1_IN_GG(x1, x2)

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

(22) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(23) Obligation:

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

EQ_LEN1_IN_GG(.(X1, Xs), .(X2, Ys)) → EQ_LEN1_IN_GG(Xs, Ys)

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

(24) PiDPToQDPProof (EQUIVALENT transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(25) Obligation:

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

EQ_LEN1_IN_GG(.(X1, Xs), .(X2, Ys)) → EQ_LEN1_IN_GG(Xs, Ys)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(26) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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_IN_GG(.(X1, Xs), .(X2, Ys)) → EQ_LEN1_IN_GG(Xs, Ys)
  The graph contains the following edges 1 > 1, 2 > 2

(27) YES