Left Termination of the query pattern subset_in_2(g, g) w.r.t. the given Prolog program could successfully be proven:

0 Prolog
  ↳1 PrologToPiTRSProof (⇐)
    ↳2 PiTRS
    ↳3 DependencyPairsProof (⇔)
    ↳4 PiDP
    ↳5 DependencyGraphProof (⇔)
    ↳6 AND
      ↳7 PiDP
        ↳8 UsableRulesProof (⇔)
        ↳9 PiDP
        ↳10 PiDPToQDPProof (⇔)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔)
        ↳13 TRUE
      ↳14 PiDP
        ↳15 UsableRulesProof (⇔)
        ↳16 PiDP
        ↳17 PiDPToQDPProof (⇐)
        ↳18 QDP
        ↳19 QDPSizeChangeProof (⇔)
        ↳20 TRUE
  ↳21 PrologToPiTRSProof (⇐)
    ↳22 PiTRS
    ↳23 DependencyPairsProof (⇔)
    ↳24 PiDP
    ↳25 DependencyGraphProof (⇔)
    ↳26 AND
      ↳27 PiDP
        ↳28 UsableRulesProof (⇔)
        ↳29 PiDP
        ↳30 PiDPToQDPProof (⇔)
        ↳31 QDP
        ↳32 QDPSizeChangeProof (⇔)
        ↳33 TRUE
      ↳34 PiDP
        ↳35 UsableRulesProof (⇔)
        ↳36 PiDP
        ↳37 PiDPToQDPProof (⇔)
        ↳38 QDP

(0) Obligation:

Clauses:

member(X, .(Y, Xs)) :- member(X, Xs).
member(X, .(X, Xs)).
subset(.(X, Xs), Ys) :- ','(member(X, Ys), subset(Xs, Ys)).
subset([], Ys).
member1(X, .(Y, Xs)) :- member1(X, Xs).
member1(X, .(X, Xs)).
subset1(.(X, Xs), Ys) :- ','(member1(X, Ys), subset1(Xs, Ys)).
subset1([], Ys).

Queries:

subset(g,g).

(1) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
subset_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:

subset_in_gg(.(X, Xs), Ys) → U2_gg(X, Xs, Ys, member_in_gg(X, Ys))
member_in_gg(X, .(Y, Xs)) → U1_gg(X, Y, Xs, member_in_gg(X, Xs))
member_in_gg(X, .(X, Xs)) → member_out_gg(X, .(X, Xs))
U1_gg(X, Y, Xs, member_out_gg(X, Xs)) → member_out_gg(X, .(Y, Xs))
U2_gg(X, Xs, Ys, member_out_gg(X, Ys)) → U3_gg(X, Xs, Ys, subset_in_gg(Xs, Ys))
subset_in_gg([], Ys) → subset_out_gg([], Ys)
U3_gg(X, Xs, Ys, subset_out_gg(Xs, Ys)) → subset_out_gg(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_gg(x1, x2)  =  subset_in_gg(x1, x2)
.(x1, x2)  =  .(x1, x2)
U2_gg(x1, x2, x3, x4)  =  U2_gg(x2, x3, x4)
member_in_gg(x1, x2)  =  member_in_gg(x1, x2)
U1_gg(x1, x2, x3, x4)  =  U1_gg(x4)
member_out_gg(x1, x2)  =  member_out_gg
U3_gg(x1, x2, x3, x4)  =  U3_gg(x4)
[]  =  []
subset_out_gg(x1, x2)  =  subset_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:

subset_in_gg(.(X, Xs), Ys) → U2_gg(X, Xs, Ys, member_in_gg(X, Ys))
member_in_gg(X, .(Y, Xs)) → U1_gg(X, Y, Xs, member_in_gg(X, Xs))
member_in_gg(X, .(X, Xs)) → member_out_gg(X, .(X, Xs))
U1_gg(X, Y, Xs, member_out_gg(X, Xs)) → member_out_gg(X, .(Y, Xs))
U2_gg(X, Xs, Ys, member_out_gg(X, Ys)) → U3_gg(X, Xs, Ys, subset_in_gg(Xs, Ys))
subset_in_gg([], Ys) → subset_out_gg([], Ys)
U3_gg(X, Xs, Ys, subset_out_gg(Xs, Ys)) → subset_out_gg(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_gg(x1, x2)  =  subset_in_gg(x1, x2)
.(x1, x2)  =  .(x1, x2)
U2_gg(x1, x2, x3, x4)  =  U2_gg(x2, x3, x4)
member_in_gg(x1, x2)  =  member_in_gg(x1, x2)
U1_gg(x1, x2, x3, x4)  =  U1_gg(x4)
member_out_gg(x1, x2)  =  member_out_gg
U3_gg(x1, x2, x3, x4)  =  U3_gg(x4)
[]  =  []
subset_out_gg(x1, x2)  =  subset_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:

SUBSET_IN_GG(.(X, Xs), Ys) → U2_GG(X, Xs, Ys, member_in_gg(X, Ys))
SUBSET_IN_GG(.(X, Xs), Ys) → MEMBER_IN_GG(X, Ys)
MEMBER_IN_GG(X, .(Y, Xs)) → U1_GG(X, Y, Xs, member_in_gg(X, Xs))
MEMBER_IN_GG(X, .(Y, Xs)) → MEMBER_IN_GG(X, Xs)
U2_GG(X, Xs, Ys, member_out_gg(X, Ys)) → U3_GG(X, Xs, Ys, subset_in_gg(Xs, Ys))
U2_GG(X, Xs, Ys, member_out_gg(X, Ys)) → SUBSET_IN_GG(Xs, Ys)

The TRS R consists of the following rules:

subset_in_gg(.(X, Xs), Ys) → U2_gg(X, Xs, Ys, member_in_gg(X, Ys))
member_in_gg(X, .(Y, Xs)) → U1_gg(X, Y, Xs, member_in_gg(X, Xs))
member_in_gg(X, .(X, Xs)) → member_out_gg(X, .(X, Xs))
U1_gg(X, Y, Xs, member_out_gg(X, Xs)) → member_out_gg(X, .(Y, Xs))
U2_gg(X, Xs, Ys, member_out_gg(X, Ys)) → U3_gg(X, Xs, Ys, subset_in_gg(Xs, Ys))
subset_in_gg([], Ys) → subset_out_gg([], Ys)
U3_gg(X, Xs, Ys, subset_out_gg(Xs, Ys)) → subset_out_gg(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_gg(x1, x2)  =  subset_in_gg(x1, x2)
.(x1, x2)  =  .(x1, x2)
U2_gg(x1, x2, x3, x4)  =  U2_gg(x2, x3, x4)
member_in_gg(x1, x2)  =  member_in_gg(x1, x2)
U1_gg(x1, x2, x3, x4)  =  U1_gg(x4)
member_out_gg(x1, x2)  =  member_out_gg
U3_gg(x1, x2, x3, x4)  =  U3_gg(x4)
[]  =  []
subset_out_gg(x1, x2)  =  subset_out_gg
SUBSET_IN_GG(x1, x2)  =  SUBSET_IN_GG(x1, x2)
U2_GG(x1, x2, x3, x4)  =  U2_GG(x2, x3, x4)
MEMBER_IN_GG(x1, x2)  =  MEMBER_IN_GG(x1, x2)
U1_GG(x1, x2, x3, x4)  =  U1_GG(x4)
U3_GG(x1, x2, x3, x4)  =  U3_GG(x4)

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

(4) Obligation:

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

SUBSET_IN_GG(.(X, Xs), Ys) → U2_GG(X, Xs, Ys, member_in_gg(X, Ys))
SUBSET_IN_GG(.(X, Xs), Ys) → MEMBER_IN_GG(X, Ys)
MEMBER_IN_GG(X, .(Y, Xs)) → U1_GG(X, Y, Xs, member_in_gg(X, Xs))
MEMBER_IN_GG(X, .(Y, Xs)) → MEMBER_IN_GG(X, Xs)
U2_GG(X, Xs, Ys, member_out_gg(X, Ys)) → U3_GG(X, Xs, Ys, subset_in_gg(Xs, Ys))
U2_GG(X, Xs, Ys, member_out_gg(X, Ys)) → SUBSET_IN_GG(Xs, Ys)

The TRS R consists of the following rules:

subset_in_gg(.(X, Xs), Ys) → U2_gg(X, Xs, Ys, member_in_gg(X, Ys))
member_in_gg(X, .(Y, Xs)) → U1_gg(X, Y, Xs, member_in_gg(X, Xs))
member_in_gg(X, .(X, Xs)) → member_out_gg(X, .(X, Xs))
U1_gg(X, Y, Xs, member_out_gg(X, Xs)) → member_out_gg(X, .(Y, Xs))
U2_gg(X, Xs, Ys, member_out_gg(X, Ys)) → U3_gg(X, Xs, Ys, subset_in_gg(Xs, Ys))
subset_in_gg([], Ys) → subset_out_gg([], Ys)
U3_gg(X, Xs, Ys, subset_out_gg(Xs, Ys)) → subset_out_gg(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_gg(x1, x2)  =  subset_in_gg(x1, x2)
.(x1, x2)  =  .(x1, x2)
U2_gg(x1, x2, x3, x4)  =  U2_gg(x2, x3, x4)
member_in_gg(x1, x2)  =  member_in_gg(x1, x2)
U1_gg(x1, x2, x3, x4)  =  U1_gg(x4)
member_out_gg(x1, x2)  =  member_out_gg
U3_gg(x1, x2, x3, x4)  =  U3_gg(x4)
[]  =  []
subset_out_gg(x1, x2)  =  subset_out_gg
SUBSET_IN_GG(x1, x2)  =  SUBSET_IN_GG(x1, x2)
U2_GG(x1, x2, x3, x4)  =  U2_GG(x2, x3, x4)
MEMBER_IN_GG(x1, x2)  =  MEMBER_IN_GG(x1, x2)
U1_GG(x1, x2, x3, x4)  =  U1_GG(x4)
U3_GG(x1, x2, x3, x4)  =  U3_GG(x4)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

MEMBER_IN_GG(X, .(Y, Xs)) → MEMBER_IN_GG(X, Xs)

The TRS R consists of the following rules:

subset_in_gg(.(X, Xs), Ys) → U2_gg(X, Xs, Ys, member_in_gg(X, Ys))
member_in_gg(X, .(Y, Xs)) → U1_gg(X, Y, Xs, member_in_gg(X, Xs))
member_in_gg(X, .(X, Xs)) → member_out_gg(X, .(X, Xs))
U1_gg(X, Y, Xs, member_out_gg(X, Xs)) → member_out_gg(X, .(Y, Xs))
U2_gg(X, Xs, Ys, member_out_gg(X, Ys)) → U3_gg(X, Xs, Ys, subset_in_gg(Xs, Ys))
subset_in_gg([], Ys) → subset_out_gg([], Ys)
U3_gg(X, Xs, Ys, subset_out_gg(Xs, Ys)) → subset_out_gg(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_gg(x1, x2)  =  subset_in_gg(x1, x2)
.(x1, x2)  =  .(x1, x2)
U2_gg(x1, x2, x3, x4)  =  U2_gg(x2, x3, x4)
member_in_gg(x1, x2)  =  member_in_gg(x1, x2)
U1_gg(x1, x2, x3, x4)  =  U1_gg(x4)
member_out_gg(x1, x2)  =  member_out_gg
U3_gg(x1, x2, x3, x4)  =  U3_gg(x4)
[]  =  []
subset_out_gg(x1, x2)  =  subset_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, .(Y, Xs)) → MEMBER_IN_GG(X, Xs)

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, .(Y, Xs)) → MEMBER_IN_GG(X, Xs)

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, .(Y, Xs)) → MEMBER_IN_GG(X, Xs)
    The graph contains the following edges 1 >= 1, 2 > 2

(13) TRUE

(14) Obligation:

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

U2_GG(X, Xs, Ys, member_out_gg(X, Ys)) → SUBSET_IN_GG(Xs, Ys)
SUBSET_IN_GG(.(X, Xs), Ys) → U2_GG(X, Xs, Ys, member_in_gg(X, Ys))

The TRS R consists of the following rules:

subset_in_gg(.(X, Xs), Ys) → U2_gg(X, Xs, Ys, member_in_gg(X, Ys))
member_in_gg(X, .(Y, Xs)) → U1_gg(X, Y, Xs, member_in_gg(X, Xs))
member_in_gg(X, .(X, Xs)) → member_out_gg(X, .(X, Xs))
U1_gg(X, Y, Xs, member_out_gg(X, Xs)) → member_out_gg(X, .(Y, Xs))
U2_gg(X, Xs, Ys, member_out_gg(X, Ys)) → U3_gg(X, Xs, Ys, subset_in_gg(Xs, Ys))
subset_in_gg([], Ys) → subset_out_gg([], Ys)
U3_gg(X, Xs, Ys, subset_out_gg(Xs, Ys)) → subset_out_gg(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_gg(x1, x2)  =  subset_in_gg(x1, x2)
.(x1, x2)  =  .(x1, x2)
U2_gg(x1, x2, x3, x4)  =  U2_gg(x2, x3, x4)
member_in_gg(x1, x2)  =  member_in_gg(x1, x2)
U1_gg(x1, x2, x3, x4)  =  U1_gg(x4)
member_out_gg(x1, x2)  =  member_out_gg
U3_gg(x1, x2, x3, x4)  =  U3_gg(x4)
[]  =  []
subset_out_gg(x1, x2)  =  subset_out_gg
SUBSET_IN_GG(x1, x2)  =  SUBSET_IN_GG(x1, x2)
U2_GG(x1, x2, x3, x4)  =  U2_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:

U2_GG(X, Xs, Ys, member_out_gg(X, Ys)) → SUBSET_IN_GG(Xs, Ys)
SUBSET_IN_GG(.(X, Xs), Ys) → U2_GG(X, Xs, Ys, member_in_gg(X, Ys))

The TRS R consists of the following rules:

member_in_gg(X, .(Y, Xs)) → U1_gg(X, Y, Xs, member_in_gg(X, Xs))
member_in_gg(X, .(X, Xs)) → member_out_gg(X, .(X, Xs))
U1_gg(X, Y, Xs, member_out_gg(X, Xs)) → member_out_gg(X, .(Y, Xs))

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
member_in_gg(x1, x2)  =  member_in_gg(x1, x2)
U1_gg(x1, x2, x3, x4)  =  U1_gg(x4)
member_out_gg(x1, x2)  =  member_out_gg
SUBSET_IN_GG(x1, x2)  =  SUBSET_IN_GG(x1, x2)
U2_GG(x1, x2, x3, x4)  =  U2_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:

U2_GG(Xs, Ys, member_out_gg) → SUBSET_IN_GG(Xs, Ys)
SUBSET_IN_GG(.(X, Xs), Ys) → U2_GG(Xs, Ys, member_in_gg(X, Ys))

The TRS R consists of the following rules:

member_in_gg(X, .(Y, Xs)) → U1_gg(member_in_gg(X, Xs))
member_in_gg(X, .(X, Xs)) → member_out_gg
U1_gg(member_out_gg) → member_out_gg

The set Q consists of the following terms:

member_in_gg(x0, x1)
U1_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:

  • SUBSET_IN_GG(.(X, Xs), Ys) → U2_GG(Xs, Ys, member_in_gg(X, Ys))
    The graph contains the following edges 1 > 1, 2 >= 2

  • U2_GG(Xs, Ys, member_out_gg) → SUBSET_IN_GG(Xs, Ys)
    The graph contains the following edges 1 >= 1, 2 >= 2

(20) TRUE

(21) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
subset_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:

subset_in_gg(.(X, Xs), Ys) → U2_gg(X, Xs, Ys, member_in_gg(X, Ys))
member_in_gg(X, .(Y, Xs)) → U1_gg(X, Y, Xs, member_in_gg(X, Xs))
member_in_gg(X, .(X, Xs)) → member_out_gg(X, .(X, Xs))
U1_gg(X, Y, Xs, member_out_gg(X, Xs)) → member_out_gg(X, .(Y, Xs))
U2_gg(X, Xs, Ys, member_out_gg(X, Ys)) → U3_gg(X, Xs, Ys, subset_in_gg(Xs, Ys))
subset_in_gg([], Ys) → subset_out_gg([], Ys)
U3_gg(X, Xs, Ys, subset_out_gg(Xs, Ys)) → subset_out_gg(.(X, Xs), Ys)

Pi is empty.

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(22) Obligation:

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

subset_in_gg(.(X, Xs), Ys) → U2_gg(X, Xs, Ys, member_in_gg(X, Ys))
member_in_gg(X, .(Y, Xs)) → U1_gg(X, Y, Xs, member_in_gg(X, Xs))
member_in_gg(X, .(X, Xs)) → member_out_gg(X, .(X, Xs))
U1_gg(X, Y, Xs, member_out_gg(X, Xs)) → member_out_gg(X, .(Y, Xs))
U2_gg(X, Xs, Ys, member_out_gg(X, Ys)) → U3_gg(X, Xs, Ys, subset_in_gg(Xs, Ys))
subset_in_gg([], Ys) → subset_out_gg([], Ys)
U3_gg(X, Xs, Ys, subset_out_gg(Xs, Ys)) → subset_out_gg(.(X, Xs), Ys)

Pi is empty.

(23) 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:

SUBSET_IN_GG(.(X, Xs), Ys) → U2_GG(X, Xs, Ys, member_in_gg(X, Ys))
SUBSET_IN_GG(.(X, Xs), Ys) → MEMBER_IN_GG(X, Ys)
MEMBER_IN_GG(X, .(Y, Xs)) → U1_GG(X, Y, Xs, member_in_gg(X, Xs))
MEMBER_IN_GG(X, .(Y, Xs)) → MEMBER_IN_GG(X, Xs)
U2_GG(X, Xs, Ys, member_out_gg(X, Ys)) → U3_GG(X, Xs, Ys, subset_in_gg(Xs, Ys))
U2_GG(X, Xs, Ys, member_out_gg(X, Ys)) → SUBSET_IN_GG(Xs, Ys)

The TRS R consists of the following rules:

subset_in_gg(.(X, Xs), Ys) → U2_gg(X, Xs, Ys, member_in_gg(X, Ys))
member_in_gg(X, .(Y, Xs)) → U1_gg(X, Y, Xs, member_in_gg(X, Xs))
member_in_gg(X, .(X, Xs)) → member_out_gg(X, .(X, Xs))
U1_gg(X, Y, Xs, member_out_gg(X, Xs)) → member_out_gg(X, .(Y, Xs))
U2_gg(X, Xs, Ys, member_out_gg(X, Ys)) → U3_gg(X, Xs, Ys, subset_in_gg(Xs, Ys))
subset_in_gg([], Ys) → subset_out_gg([], Ys)
U3_gg(X, Xs, Ys, subset_out_gg(Xs, Ys)) → subset_out_gg(.(X, Xs), Ys)

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

(24) Obligation:

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

SUBSET_IN_GG(.(X, Xs), Ys) → U2_GG(X, Xs, Ys, member_in_gg(X, Ys))
SUBSET_IN_GG(.(X, Xs), Ys) → MEMBER_IN_GG(X, Ys)
MEMBER_IN_GG(X, .(Y, Xs)) → U1_GG(X, Y, Xs, member_in_gg(X, Xs))
MEMBER_IN_GG(X, .(Y, Xs)) → MEMBER_IN_GG(X, Xs)
U2_GG(X, Xs, Ys, member_out_gg(X, Ys)) → U3_GG(X, Xs, Ys, subset_in_gg(Xs, Ys))
U2_GG(X, Xs, Ys, member_out_gg(X, Ys)) → SUBSET_IN_GG(Xs, Ys)

The TRS R consists of the following rules:

subset_in_gg(.(X, Xs), Ys) → U2_gg(X, Xs, Ys, member_in_gg(X, Ys))
member_in_gg(X, .(Y, Xs)) → U1_gg(X, Y, Xs, member_in_gg(X, Xs))
member_in_gg(X, .(X, Xs)) → member_out_gg(X, .(X, Xs))
U1_gg(X, Y, Xs, member_out_gg(X, Xs)) → member_out_gg(X, .(Y, Xs))
U2_gg(X, Xs, Ys, member_out_gg(X, Ys)) → U3_gg(X, Xs, Ys, subset_in_gg(Xs, Ys))
subset_in_gg([], Ys) → subset_out_gg([], Ys)
U3_gg(X, Xs, Ys, subset_out_gg(Xs, Ys)) → subset_out_gg(.(X, Xs), Ys)

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

(25) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 3 less nodes.

(26) Complex Obligation (AND)

(27) Obligation:

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

MEMBER_IN_GG(X, .(Y, Xs)) → MEMBER_IN_GG(X, Xs)

The TRS R consists of the following rules:

subset_in_gg(.(X, Xs), Ys) → U2_gg(X, Xs, Ys, member_in_gg(X, Ys))
member_in_gg(X, .(Y, Xs)) → U1_gg(X, Y, Xs, member_in_gg(X, Xs))
member_in_gg(X, .(X, Xs)) → member_out_gg(X, .(X, Xs))
U1_gg(X, Y, Xs, member_out_gg(X, Xs)) → member_out_gg(X, .(Y, Xs))
U2_gg(X, Xs, Ys, member_out_gg(X, Ys)) → U3_gg(X, Xs, Ys, subset_in_gg(Xs, Ys))
subset_in_gg([], Ys) → subset_out_gg([], Ys)
U3_gg(X, Xs, Ys, subset_out_gg(Xs, Ys)) → subset_out_gg(.(X, Xs), Ys)

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

(28) UsableRulesProof (EQUIVALENT transformation)

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

(29) Obligation:

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

MEMBER_IN_GG(X, .(Y, Xs)) → MEMBER_IN_GG(X, Xs)

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

(30) PiDPToQDPProof (EQUIVALENT transformation)

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

(31) Obligation:

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

MEMBER_IN_GG(X, .(Y, Xs)) → MEMBER_IN_GG(X, Xs)

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

(32) 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, .(Y, Xs)) → MEMBER_IN_GG(X, Xs)
    The graph contains the following edges 1 >= 1, 2 > 2

(33) TRUE

(34) Obligation:

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

U2_GG(X, Xs, Ys, member_out_gg(X, Ys)) → SUBSET_IN_GG(Xs, Ys)
SUBSET_IN_GG(.(X, Xs), Ys) → U2_GG(X, Xs, Ys, member_in_gg(X, Ys))

The TRS R consists of the following rules:

subset_in_gg(.(X, Xs), Ys) → U2_gg(X, Xs, Ys, member_in_gg(X, Ys))
member_in_gg(X, .(Y, Xs)) → U1_gg(X, Y, Xs, member_in_gg(X, Xs))
member_in_gg(X, .(X, Xs)) → member_out_gg(X, .(X, Xs))
U1_gg(X, Y, Xs, member_out_gg(X, Xs)) → member_out_gg(X, .(Y, Xs))
U2_gg(X, Xs, Ys, member_out_gg(X, Ys)) → U3_gg(X, Xs, Ys, subset_in_gg(Xs, Ys))
subset_in_gg([], Ys) → subset_out_gg([], Ys)
U3_gg(X, Xs, Ys, subset_out_gg(Xs, Ys)) → subset_out_gg(.(X, Xs), Ys)

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

(35) UsableRulesProof (EQUIVALENT transformation)

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

(36) Obligation:

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

U2_GG(X, Xs, Ys, member_out_gg(X, Ys)) → SUBSET_IN_GG(Xs, Ys)
SUBSET_IN_GG(.(X, Xs), Ys) → U2_GG(X, Xs, Ys, member_in_gg(X, Ys))

The TRS R consists of the following rules:

member_in_gg(X, .(Y, Xs)) → U1_gg(X, Y, Xs, member_in_gg(X, Xs))
member_in_gg(X, .(X, Xs)) → member_out_gg(X, .(X, Xs))
U1_gg(X, Y, Xs, member_out_gg(X, Xs)) → member_out_gg(X, .(Y, Xs))

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

(37) PiDPToQDPProof (EQUIVALENT transformation)

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

(38) Obligation:

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

U2_GG(X, Xs, Ys, member_out_gg(X, Ys)) → SUBSET_IN_GG(Xs, Ys)
SUBSET_IN_GG(.(X, Xs), Ys) → U2_GG(X, Xs, Ys, member_in_gg(X, Ys))

The TRS R consists of the following rules:

member_in_gg(X, .(Y, Xs)) → U1_gg(X, Y, Xs, member_in_gg(X, Xs))
member_in_gg(X, .(X, Xs)) → member_out_gg(X, .(X, Xs))
U1_gg(X, Y, Xs, member_out_gg(X, Xs)) → member_out_gg(X, .(Y, Xs))

The set Q consists of the following terms:

member_in_gg(x0, x1)
U1_gg(x0, x1, x2, x3)

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