(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).
Query: subset(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:
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) YES
(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) YES