(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).
Queries:
subset(a,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: (f,b)
member_in: (f,b)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
subset_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, member_in_ag(X, Ys))
member_in_ag(X, .(Y, Xs)) → U1_ag(X, Y, Xs, member_in_ag(X, Xs))
member_in_ag(X, .(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(X, Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))
U2_ag(X, Xs, Ys, member_out_ag(X, Ys)) → U3_ag(X, Xs, Ys, subset_in_ag(Xs, Ys))
subset_in_ag([], Ys) → subset_out_ag([], Ys)
U3_ag(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ag(.(X, Xs), Ys)
The argument filtering Pi contains the following mapping:
subset_in_ag(
x1,
x2) =
subset_in_ag(
x2)
U2_ag(
x1,
x2,
x3,
x4) =
U2_ag(
x3,
x4)
member_in_ag(
x1,
x2) =
member_in_ag(
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ag(
x1,
x2,
x3,
x4) =
U1_ag(
x4)
member_out_ag(
x1,
x2) =
member_out_ag(
x1)
U3_ag(
x1,
x2,
x3,
x4) =
U3_ag(
x1,
x4)
subset_out_ag(
x1,
x2) =
subset_out_ag(
x1)
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_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, member_in_ag(X, Ys))
member_in_ag(X, .(Y, Xs)) → U1_ag(X, Y, Xs, member_in_ag(X, Xs))
member_in_ag(X, .(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(X, Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))
U2_ag(X, Xs, Ys, member_out_ag(X, Ys)) → U3_ag(X, Xs, Ys, subset_in_ag(Xs, Ys))
subset_in_ag([], Ys) → subset_out_ag([], Ys)
U3_ag(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ag(.(X, Xs), Ys)
The argument filtering Pi contains the following mapping:
subset_in_ag(
x1,
x2) =
subset_in_ag(
x2)
U2_ag(
x1,
x2,
x3,
x4) =
U2_ag(
x3,
x4)
member_in_ag(
x1,
x2) =
member_in_ag(
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ag(
x1,
x2,
x3,
x4) =
U1_ag(
x4)
member_out_ag(
x1,
x2) =
member_out_ag(
x1)
U3_ag(
x1,
x2,
x3,
x4) =
U3_ag(
x1,
x4)
subset_out_ag(
x1,
x2) =
subset_out_ag(
x1)
(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_AG(.(X, Xs), Ys) → U2_AG(X, Xs, Ys, member_in_ag(X, Ys))
SUBSET_IN_AG(.(X, Xs), Ys) → MEMBER_IN_AG(X, Ys)
MEMBER_IN_AG(X, .(Y, Xs)) → U1_AG(X, Y, Xs, member_in_ag(X, Xs))
MEMBER_IN_AG(X, .(Y, Xs)) → MEMBER_IN_AG(X, Xs)
U2_AG(X, Xs, Ys, member_out_ag(X, Ys)) → U3_AG(X, Xs, Ys, subset_in_ag(Xs, Ys))
U2_AG(X, Xs, Ys, member_out_ag(X, Ys)) → SUBSET_IN_AG(Xs, Ys)
The TRS R consists of the following rules:
subset_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, member_in_ag(X, Ys))
member_in_ag(X, .(Y, Xs)) → U1_ag(X, Y, Xs, member_in_ag(X, Xs))
member_in_ag(X, .(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(X, Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))
U2_ag(X, Xs, Ys, member_out_ag(X, Ys)) → U3_ag(X, Xs, Ys, subset_in_ag(Xs, Ys))
subset_in_ag([], Ys) → subset_out_ag([], Ys)
U3_ag(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ag(.(X, Xs), Ys)
The argument filtering Pi contains the following mapping:
subset_in_ag(
x1,
x2) =
subset_in_ag(
x2)
U2_ag(
x1,
x2,
x3,
x4) =
U2_ag(
x3,
x4)
member_in_ag(
x1,
x2) =
member_in_ag(
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ag(
x1,
x2,
x3,
x4) =
U1_ag(
x4)
member_out_ag(
x1,
x2) =
member_out_ag(
x1)
U3_ag(
x1,
x2,
x3,
x4) =
U3_ag(
x1,
x4)
subset_out_ag(
x1,
x2) =
subset_out_ag(
x1)
SUBSET_IN_AG(
x1,
x2) =
SUBSET_IN_AG(
x2)
U2_AG(
x1,
x2,
x3,
x4) =
U2_AG(
x3,
x4)
MEMBER_IN_AG(
x1,
x2) =
MEMBER_IN_AG(
x2)
U1_AG(
x1,
x2,
x3,
x4) =
U1_AG(
x4)
U3_AG(
x1,
x2,
x3,
x4) =
U3_AG(
x1,
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_AG(.(X, Xs), Ys) → U2_AG(X, Xs, Ys, member_in_ag(X, Ys))
SUBSET_IN_AG(.(X, Xs), Ys) → MEMBER_IN_AG(X, Ys)
MEMBER_IN_AG(X, .(Y, Xs)) → U1_AG(X, Y, Xs, member_in_ag(X, Xs))
MEMBER_IN_AG(X, .(Y, Xs)) → MEMBER_IN_AG(X, Xs)
U2_AG(X, Xs, Ys, member_out_ag(X, Ys)) → U3_AG(X, Xs, Ys, subset_in_ag(Xs, Ys))
U2_AG(X, Xs, Ys, member_out_ag(X, Ys)) → SUBSET_IN_AG(Xs, Ys)
The TRS R consists of the following rules:
subset_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, member_in_ag(X, Ys))
member_in_ag(X, .(Y, Xs)) → U1_ag(X, Y, Xs, member_in_ag(X, Xs))
member_in_ag(X, .(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(X, Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))
U2_ag(X, Xs, Ys, member_out_ag(X, Ys)) → U3_ag(X, Xs, Ys, subset_in_ag(Xs, Ys))
subset_in_ag([], Ys) → subset_out_ag([], Ys)
U3_ag(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ag(.(X, Xs), Ys)
The argument filtering Pi contains the following mapping:
subset_in_ag(
x1,
x2) =
subset_in_ag(
x2)
U2_ag(
x1,
x2,
x3,
x4) =
U2_ag(
x3,
x4)
member_in_ag(
x1,
x2) =
member_in_ag(
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ag(
x1,
x2,
x3,
x4) =
U1_ag(
x4)
member_out_ag(
x1,
x2) =
member_out_ag(
x1)
U3_ag(
x1,
x2,
x3,
x4) =
U3_ag(
x1,
x4)
subset_out_ag(
x1,
x2) =
subset_out_ag(
x1)
SUBSET_IN_AG(
x1,
x2) =
SUBSET_IN_AG(
x2)
U2_AG(
x1,
x2,
x3,
x4) =
U2_AG(
x3,
x4)
MEMBER_IN_AG(
x1,
x2) =
MEMBER_IN_AG(
x2)
U1_AG(
x1,
x2,
x3,
x4) =
U1_AG(
x4)
U3_AG(
x1,
x2,
x3,
x4) =
U3_AG(
x1,
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_AG(X, .(Y, Xs)) → MEMBER_IN_AG(X, Xs)
The TRS R consists of the following rules:
subset_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, member_in_ag(X, Ys))
member_in_ag(X, .(Y, Xs)) → U1_ag(X, Y, Xs, member_in_ag(X, Xs))
member_in_ag(X, .(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(X, Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))
U2_ag(X, Xs, Ys, member_out_ag(X, Ys)) → U3_ag(X, Xs, Ys, subset_in_ag(Xs, Ys))
subset_in_ag([], Ys) → subset_out_ag([], Ys)
U3_ag(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ag(.(X, Xs), Ys)
The argument filtering Pi contains the following mapping:
subset_in_ag(
x1,
x2) =
subset_in_ag(
x2)
U2_ag(
x1,
x2,
x3,
x4) =
U2_ag(
x3,
x4)
member_in_ag(
x1,
x2) =
member_in_ag(
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ag(
x1,
x2,
x3,
x4) =
U1_ag(
x4)
member_out_ag(
x1,
x2) =
member_out_ag(
x1)
U3_ag(
x1,
x2,
x3,
x4) =
U3_ag(
x1,
x4)
subset_out_ag(
x1,
x2) =
subset_out_ag(
x1)
MEMBER_IN_AG(
x1,
x2) =
MEMBER_IN_AG(
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_AG(X, .(Y, Xs)) → MEMBER_IN_AG(X, Xs)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
MEMBER_IN_AG(
x1,
x2) =
MEMBER_IN_AG(
x2)
We have to consider all (P,R,Pi)-chains
(10) PiDPToQDPProof (SOUND 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_AG(.(Y, Xs)) → MEMBER_IN_AG(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_AG(.(Y, Xs)) → MEMBER_IN_AG(Xs)
The graph contains the following edges 1 > 1
(13) TRUE
(14) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
U2_AG(X, Xs, Ys, member_out_ag(X, Ys)) → SUBSET_IN_AG(Xs, Ys)
SUBSET_IN_AG(.(X, Xs), Ys) → U2_AG(X, Xs, Ys, member_in_ag(X, Ys))
The TRS R consists of the following rules:
subset_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, member_in_ag(X, Ys))
member_in_ag(X, .(Y, Xs)) → U1_ag(X, Y, Xs, member_in_ag(X, Xs))
member_in_ag(X, .(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(X, Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))
U2_ag(X, Xs, Ys, member_out_ag(X, Ys)) → U3_ag(X, Xs, Ys, subset_in_ag(Xs, Ys))
subset_in_ag([], Ys) → subset_out_ag([], Ys)
U3_ag(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ag(.(X, Xs), Ys)
The argument filtering Pi contains the following mapping:
subset_in_ag(
x1,
x2) =
subset_in_ag(
x2)
U2_ag(
x1,
x2,
x3,
x4) =
U2_ag(
x3,
x4)
member_in_ag(
x1,
x2) =
member_in_ag(
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ag(
x1,
x2,
x3,
x4) =
U1_ag(
x4)
member_out_ag(
x1,
x2) =
member_out_ag(
x1)
U3_ag(
x1,
x2,
x3,
x4) =
U3_ag(
x1,
x4)
subset_out_ag(
x1,
x2) =
subset_out_ag(
x1)
SUBSET_IN_AG(
x1,
x2) =
SUBSET_IN_AG(
x2)
U2_AG(
x1,
x2,
x3,
x4) =
U2_AG(
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_AG(X, Xs, Ys, member_out_ag(X, Ys)) → SUBSET_IN_AG(Xs, Ys)
SUBSET_IN_AG(.(X, Xs), Ys) → U2_AG(X, Xs, Ys, member_in_ag(X, Ys))
The TRS R consists of the following rules:
member_in_ag(X, .(Y, Xs)) → U1_ag(X, Y, Xs, member_in_ag(X, Xs))
member_in_ag(X, .(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(X, Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))
The argument filtering Pi contains the following mapping:
member_in_ag(
x1,
x2) =
member_in_ag(
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ag(
x1,
x2,
x3,
x4) =
U1_ag(
x4)
member_out_ag(
x1,
x2) =
member_out_ag(
x1)
SUBSET_IN_AG(
x1,
x2) =
SUBSET_IN_AG(
x2)
U2_AG(
x1,
x2,
x3,
x4) =
U2_AG(
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_AG(Ys, member_out_ag(X)) → SUBSET_IN_AG(Ys)
SUBSET_IN_AG(Ys) → U2_AG(Ys, member_in_ag(Ys))
The TRS R consists of the following rules:
member_in_ag(.(Y, Xs)) → U1_ag(member_in_ag(Xs))
member_in_ag(.(X, Xs)) → member_out_ag(X)
U1_ag(member_out_ag(X)) → member_out_ag(X)
The set Q consists of the following terms:
member_in_ag(x0)
U1_ag(x0)
We have to consider all (P,Q,R)-chains.
(19) Narrowing (SOUND transformation)
By narrowing [LPAR04] the rule
SUBSET_IN_AG(
Ys) →
U2_AG(
Ys,
member_in_ag(
Ys)) at position [1] we obtained the following new rules [LPAR04]:
SUBSET_IN_AG(.(x0, x1)) → U2_AG(.(x0, x1), U1_ag(member_in_ag(x1)))
SUBSET_IN_AG(.(x0, x1)) → U2_AG(.(x0, x1), member_out_ag(x0))
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
U2_AG(Ys, member_out_ag(X)) → SUBSET_IN_AG(Ys)
SUBSET_IN_AG(.(x0, x1)) → U2_AG(.(x0, x1), U1_ag(member_in_ag(x1)))
SUBSET_IN_AG(.(x0, x1)) → U2_AG(.(x0, x1), member_out_ag(x0))
The TRS R consists of the following rules:
member_in_ag(.(Y, Xs)) → U1_ag(member_in_ag(Xs))
member_in_ag(.(X, Xs)) → member_out_ag(X)
U1_ag(member_out_ag(X)) → member_out_ag(X)
The set Q consists of the following terms:
member_in_ag(x0)
U1_ag(x0)
We have to consider all (P,Q,R)-chains.
(21) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
U2_AG(
Ys,
member_out_ag(
X)) →
SUBSET_IN_AG(
Ys) we obtained the following new rules [LPAR04]:
U2_AG(.(z0, z1), member_out_ag(x1)) → SUBSET_IN_AG(.(z0, z1))
U2_AG(.(z0, z1), member_out_ag(z0)) → SUBSET_IN_AG(.(z0, z1))
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
SUBSET_IN_AG(.(x0, x1)) → U2_AG(.(x0, x1), U1_ag(member_in_ag(x1)))
SUBSET_IN_AG(.(x0, x1)) → U2_AG(.(x0, x1), member_out_ag(x0))
U2_AG(.(z0, z1), member_out_ag(x1)) → SUBSET_IN_AG(.(z0, z1))
U2_AG(.(z0, z1), member_out_ag(z0)) → SUBSET_IN_AG(.(z0, z1))
The TRS R consists of the following rules:
member_in_ag(.(Y, Xs)) → U1_ag(member_in_ag(Xs))
member_in_ag(.(X, Xs)) → member_out_ag(X)
U1_ag(member_out_ag(X)) → member_out_ag(X)
The set Q consists of the following terms:
member_in_ag(x0)
U1_ag(x0)
We have to consider all (P,Q,R)-chains.
(23) NonTerminationProof (EQUIVALENT transformation)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:
s =
U2_AG(
.(
z0,
z1),
member_out_ag(
x1')) evaluates to t =
U2_AG(
.(
z0,
z1),
member_out_ag(
z0))
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [x1' / z0]
- Semiunifier: [ ]
Rewriting sequenceU2_AG(.(z0, z1), member_out_ag(x1')) →
SUBSET_IN_AG(
.(
z0,
z1))
with rule
U2_AG(
.(
z0',
z1'),
member_out_ag(
x1'')) →
SUBSET_IN_AG(
.(
z0',
z1')) at position [] and matcher [
z0' /
z0,
z1' /
z1,
x1'' /
x1']
SUBSET_IN_AG(.(z0, z1)) →
U2_AG(
.(
z0,
z1),
member_out_ag(
z0))
with rule
SUBSET_IN_AG(
.(
x0,
x1)) →
U2_AG(
.(
x0,
x1),
member_out_ag(
x0))
Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence
All these steps are and every following step will be a correct step w.r.t to Q.
(24) FALSE
(25) 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: (f,b)
member_in: (f,b)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
subset_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, member_in_ag(X, Ys))
member_in_ag(X, .(Y, Xs)) → U1_ag(X, Y, Xs, member_in_ag(X, Xs))
member_in_ag(X, .(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(X, Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))
U2_ag(X, Xs, Ys, member_out_ag(X, Ys)) → U3_ag(X, Xs, Ys, subset_in_ag(Xs, Ys))
subset_in_ag([], Ys) → subset_out_ag([], Ys)
U3_ag(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ag(.(X, Xs), Ys)
The argument filtering Pi contains the following mapping:
subset_in_ag(
x1,
x2) =
subset_in_ag(
x2)
U2_ag(
x1,
x2,
x3,
x4) =
U2_ag(
x3,
x4)
member_in_ag(
x1,
x2) =
member_in_ag(
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ag(
x1,
x2,
x3,
x4) =
U1_ag(
x2,
x3,
x4)
member_out_ag(
x1,
x2) =
member_out_ag(
x1,
x2)
U3_ag(
x1,
x2,
x3,
x4) =
U3_ag(
x1,
x3,
x4)
subset_out_ag(
x1,
x2) =
subset_out_ag(
x1,
x2)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(26) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
subset_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, member_in_ag(X, Ys))
member_in_ag(X, .(Y, Xs)) → U1_ag(X, Y, Xs, member_in_ag(X, Xs))
member_in_ag(X, .(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(X, Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))
U2_ag(X, Xs, Ys, member_out_ag(X, Ys)) → U3_ag(X, Xs, Ys, subset_in_ag(Xs, Ys))
subset_in_ag([], Ys) → subset_out_ag([], Ys)
U3_ag(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ag(.(X, Xs), Ys)
The argument filtering Pi contains the following mapping:
subset_in_ag(
x1,
x2) =
subset_in_ag(
x2)
U2_ag(
x1,
x2,
x3,
x4) =
U2_ag(
x3,
x4)
member_in_ag(
x1,
x2) =
member_in_ag(
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ag(
x1,
x2,
x3,
x4) =
U1_ag(
x2,
x3,
x4)
member_out_ag(
x1,
x2) =
member_out_ag(
x1,
x2)
U3_ag(
x1,
x2,
x3,
x4) =
U3_ag(
x1,
x3,
x4)
subset_out_ag(
x1,
x2) =
subset_out_ag(
x1,
x2)
(27) 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_AG(.(X, Xs), Ys) → U2_AG(X, Xs, Ys, member_in_ag(X, Ys))
SUBSET_IN_AG(.(X, Xs), Ys) → MEMBER_IN_AG(X, Ys)
MEMBER_IN_AG(X, .(Y, Xs)) → U1_AG(X, Y, Xs, member_in_ag(X, Xs))
MEMBER_IN_AG(X, .(Y, Xs)) → MEMBER_IN_AG(X, Xs)
U2_AG(X, Xs, Ys, member_out_ag(X, Ys)) → U3_AG(X, Xs, Ys, subset_in_ag(Xs, Ys))
U2_AG(X, Xs, Ys, member_out_ag(X, Ys)) → SUBSET_IN_AG(Xs, Ys)
The TRS R consists of the following rules:
subset_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, member_in_ag(X, Ys))
member_in_ag(X, .(Y, Xs)) → U1_ag(X, Y, Xs, member_in_ag(X, Xs))
member_in_ag(X, .(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(X, Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))
U2_ag(X, Xs, Ys, member_out_ag(X, Ys)) → U3_ag(X, Xs, Ys, subset_in_ag(Xs, Ys))
subset_in_ag([], Ys) → subset_out_ag([], Ys)
U3_ag(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ag(.(X, Xs), Ys)
The argument filtering Pi contains the following mapping:
subset_in_ag(
x1,
x2) =
subset_in_ag(
x2)
U2_ag(
x1,
x2,
x3,
x4) =
U2_ag(
x3,
x4)
member_in_ag(
x1,
x2) =
member_in_ag(
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ag(
x1,
x2,
x3,
x4) =
U1_ag(
x2,
x3,
x4)
member_out_ag(
x1,
x2) =
member_out_ag(
x1,
x2)
U3_ag(
x1,
x2,
x3,
x4) =
U3_ag(
x1,
x3,
x4)
subset_out_ag(
x1,
x2) =
subset_out_ag(
x1,
x2)
SUBSET_IN_AG(
x1,
x2) =
SUBSET_IN_AG(
x2)
U2_AG(
x1,
x2,
x3,
x4) =
U2_AG(
x3,
x4)
MEMBER_IN_AG(
x1,
x2) =
MEMBER_IN_AG(
x2)
U1_AG(
x1,
x2,
x3,
x4) =
U1_AG(
x2,
x3,
x4)
U3_AG(
x1,
x2,
x3,
x4) =
U3_AG(
x1,
x3,
x4)
We have to consider all (P,R,Pi)-chains
(28) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
SUBSET_IN_AG(.(X, Xs), Ys) → U2_AG(X, Xs, Ys, member_in_ag(X, Ys))
SUBSET_IN_AG(.(X, Xs), Ys) → MEMBER_IN_AG(X, Ys)
MEMBER_IN_AG(X, .(Y, Xs)) → U1_AG(X, Y, Xs, member_in_ag(X, Xs))
MEMBER_IN_AG(X, .(Y, Xs)) → MEMBER_IN_AG(X, Xs)
U2_AG(X, Xs, Ys, member_out_ag(X, Ys)) → U3_AG(X, Xs, Ys, subset_in_ag(Xs, Ys))
U2_AG(X, Xs, Ys, member_out_ag(X, Ys)) → SUBSET_IN_AG(Xs, Ys)
The TRS R consists of the following rules:
subset_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, member_in_ag(X, Ys))
member_in_ag(X, .(Y, Xs)) → U1_ag(X, Y, Xs, member_in_ag(X, Xs))
member_in_ag(X, .(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(X, Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))
U2_ag(X, Xs, Ys, member_out_ag(X, Ys)) → U3_ag(X, Xs, Ys, subset_in_ag(Xs, Ys))
subset_in_ag([], Ys) → subset_out_ag([], Ys)
U3_ag(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ag(.(X, Xs), Ys)
The argument filtering Pi contains the following mapping:
subset_in_ag(
x1,
x2) =
subset_in_ag(
x2)
U2_ag(
x1,
x2,
x3,
x4) =
U2_ag(
x3,
x4)
member_in_ag(
x1,
x2) =
member_in_ag(
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ag(
x1,
x2,
x3,
x4) =
U1_ag(
x2,
x3,
x4)
member_out_ag(
x1,
x2) =
member_out_ag(
x1,
x2)
U3_ag(
x1,
x2,
x3,
x4) =
U3_ag(
x1,
x3,
x4)
subset_out_ag(
x1,
x2) =
subset_out_ag(
x1,
x2)
SUBSET_IN_AG(
x1,
x2) =
SUBSET_IN_AG(
x2)
U2_AG(
x1,
x2,
x3,
x4) =
U2_AG(
x3,
x4)
MEMBER_IN_AG(
x1,
x2) =
MEMBER_IN_AG(
x2)
U1_AG(
x1,
x2,
x3,
x4) =
U1_AG(
x2,
x3,
x4)
U3_AG(
x1,
x2,
x3,
x4) =
U3_AG(
x1,
x3,
x4)
We have to consider all (P,R,Pi)-chains
(29) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 3 less nodes.
(30) Complex Obligation (AND)
(31) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
MEMBER_IN_AG(X, .(Y, Xs)) → MEMBER_IN_AG(X, Xs)
The TRS R consists of the following rules:
subset_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, member_in_ag(X, Ys))
member_in_ag(X, .(Y, Xs)) → U1_ag(X, Y, Xs, member_in_ag(X, Xs))
member_in_ag(X, .(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(X, Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))
U2_ag(X, Xs, Ys, member_out_ag(X, Ys)) → U3_ag(X, Xs, Ys, subset_in_ag(Xs, Ys))
subset_in_ag([], Ys) → subset_out_ag([], Ys)
U3_ag(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ag(.(X, Xs), Ys)
The argument filtering Pi contains the following mapping:
subset_in_ag(
x1,
x2) =
subset_in_ag(
x2)
U2_ag(
x1,
x2,
x3,
x4) =
U2_ag(
x3,
x4)
member_in_ag(
x1,
x2) =
member_in_ag(
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ag(
x1,
x2,
x3,
x4) =
U1_ag(
x2,
x3,
x4)
member_out_ag(
x1,
x2) =
member_out_ag(
x1,
x2)
U3_ag(
x1,
x2,
x3,
x4) =
U3_ag(
x1,
x3,
x4)
subset_out_ag(
x1,
x2) =
subset_out_ag(
x1,
x2)
MEMBER_IN_AG(
x1,
x2) =
MEMBER_IN_AG(
x2)
We have to consider all (P,R,Pi)-chains
(32) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(33) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
MEMBER_IN_AG(X, .(Y, Xs)) → MEMBER_IN_AG(X, Xs)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
MEMBER_IN_AG(
x1,
x2) =
MEMBER_IN_AG(
x2)
We have to consider all (P,R,Pi)-chains
(34) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(35) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MEMBER_IN_AG(.(Y, Xs)) → MEMBER_IN_AG(Xs)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(36) 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_AG(.(Y, Xs)) → MEMBER_IN_AG(Xs)
The graph contains the following edges 1 > 1
(37) TRUE
(38) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
U2_AG(X, Xs, Ys, member_out_ag(X, Ys)) → SUBSET_IN_AG(Xs, Ys)
SUBSET_IN_AG(.(X, Xs), Ys) → U2_AG(X, Xs, Ys, member_in_ag(X, Ys))
The TRS R consists of the following rules:
subset_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, member_in_ag(X, Ys))
member_in_ag(X, .(Y, Xs)) → U1_ag(X, Y, Xs, member_in_ag(X, Xs))
member_in_ag(X, .(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(X, Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))
U2_ag(X, Xs, Ys, member_out_ag(X, Ys)) → U3_ag(X, Xs, Ys, subset_in_ag(Xs, Ys))
subset_in_ag([], Ys) → subset_out_ag([], Ys)
U3_ag(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ag(.(X, Xs), Ys)
The argument filtering Pi contains the following mapping:
subset_in_ag(
x1,
x2) =
subset_in_ag(
x2)
U2_ag(
x1,
x2,
x3,
x4) =
U2_ag(
x3,
x4)
member_in_ag(
x1,
x2) =
member_in_ag(
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ag(
x1,
x2,
x3,
x4) =
U1_ag(
x2,
x3,
x4)
member_out_ag(
x1,
x2) =
member_out_ag(
x1,
x2)
U3_ag(
x1,
x2,
x3,
x4) =
U3_ag(
x1,
x3,
x4)
subset_out_ag(
x1,
x2) =
subset_out_ag(
x1,
x2)
SUBSET_IN_AG(
x1,
x2) =
SUBSET_IN_AG(
x2)
U2_AG(
x1,
x2,
x3,
x4) =
U2_AG(
x3,
x4)
We have to consider all (P,R,Pi)-chains
(39) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(40) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
U2_AG(X, Xs, Ys, member_out_ag(X, Ys)) → SUBSET_IN_AG(Xs, Ys)
SUBSET_IN_AG(.(X, Xs), Ys) → U2_AG(X, Xs, Ys, member_in_ag(X, Ys))
The TRS R consists of the following rules:
member_in_ag(X, .(Y, Xs)) → U1_ag(X, Y, Xs, member_in_ag(X, Xs))
member_in_ag(X, .(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(X, Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))
The argument filtering Pi contains the following mapping:
member_in_ag(
x1,
x2) =
member_in_ag(
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ag(
x1,
x2,
x3,
x4) =
U1_ag(
x2,
x3,
x4)
member_out_ag(
x1,
x2) =
member_out_ag(
x1,
x2)
SUBSET_IN_AG(
x1,
x2) =
SUBSET_IN_AG(
x2)
U2_AG(
x1,
x2,
x3,
x4) =
U2_AG(
x3,
x4)
We have to consider all (P,R,Pi)-chains
(41) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(42) Obligation:
Q DP problem:
The TRS P consists of the following rules:
U2_AG(Ys, member_out_ag(X, Ys)) → SUBSET_IN_AG(Ys)
SUBSET_IN_AG(Ys) → U2_AG(Ys, member_in_ag(Ys))
The TRS R consists of the following rules:
member_in_ag(.(Y, Xs)) → U1_ag(Y, Xs, member_in_ag(Xs))
member_in_ag(.(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))
The set Q consists of the following terms:
member_in_ag(x0)
U1_ag(x0, x1, x2)
We have to consider all (P,Q,R)-chains.
(43) Narrowing (SOUND transformation)
By narrowing [LPAR04] the rule
SUBSET_IN_AG(
Ys) →
U2_AG(
Ys,
member_in_ag(
Ys)) at position [1] we obtained the following new rules [LPAR04]:
SUBSET_IN_AG(.(x0, x1)) → U2_AG(.(x0, x1), U1_ag(x0, x1, member_in_ag(x1)))
SUBSET_IN_AG(.(x0, x1)) → U2_AG(.(x0, x1), member_out_ag(x0, .(x0, x1)))
(44) Obligation:
Q DP problem:
The TRS P consists of the following rules:
U2_AG(Ys, member_out_ag(X, Ys)) → SUBSET_IN_AG(Ys)
SUBSET_IN_AG(.(x0, x1)) → U2_AG(.(x0, x1), U1_ag(x0, x1, member_in_ag(x1)))
SUBSET_IN_AG(.(x0, x1)) → U2_AG(.(x0, x1), member_out_ag(x0, .(x0, x1)))
The TRS R consists of the following rules:
member_in_ag(.(Y, Xs)) → U1_ag(Y, Xs, member_in_ag(Xs))
member_in_ag(.(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))
The set Q consists of the following terms:
member_in_ag(x0)
U1_ag(x0, x1, x2)
We have to consider all (P,Q,R)-chains.
(45) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
U2_AG(
Ys,
member_out_ag(
X,
Ys)) →
SUBSET_IN_AG(
Ys) we obtained the following new rules [LPAR04]:
U2_AG(.(z0, z1), member_out_ag(x1, .(z0, z1))) → SUBSET_IN_AG(.(z0, z1))
U2_AG(.(z0, z1), member_out_ag(z0, .(z0, z1))) → SUBSET_IN_AG(.(z0, z1))
(46) Obligation:
Q DP problem:
The TRS P consists of the following rules:
SUBSET_IN_AG(.(x0, x1)) → U2_AG(.(x0, x1), U1_ag(x0, x1, member_in_ag(x1)))
SUBSET_IN_AG(.(x0, x1)) → U2_AG(.(x0, x1), member_out_ag(x0, .(x0, x1)))
U2_AG(.(z0, z1), member_out_ag(x1, .(z0, z1))) → SUBSET_IN_AG(.(z0, z1))
U2_AG(.(z0, z1), member_out_ag(z0, .(z0, z1))) → SUBSET_IN_AG(.(z0, z1))
The TRS R consists of the following rules:
member_in_ag(.(Y, Xs)) → U1_ag(Y, Xs, member_in_ag(Xs))
member_in_ag(.(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))
The set Q consists of the following terms:
member_in_ag(x0)
U1_ag(x0, x1, x2)
We have to consider all (P,Q,R)-chains.
(47) NonTerminationProof (EQUIVALENT transformation)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:
s =
U2_AG(
.(
z0,
z1),
member_out_ag(
x1',
.(
z0,
z1))) evaluates to t =
U2_AG(
.(
z0,
z1),
member_out_ag(
z0,
.(
z0,
z1)))
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [x1' / z0]
Rewriting sequenceU2_AG(.(z0, z1), member_out_ag(x1', .(z0, z1))) →
SUBSET_IN_AG(
.(
z0,
z1))
with rule
U2_AG(
.(
z0',
z1'),
member_out_ag(
x1'',
.(
z0',
z1'))) →
SUBSET_IN_AG(
.(
z0',
z1')) at position [] and matcher [
z0' /
z0,
z1' /
z1,
x1'' /
x1']
SUBSET_IN_AG(.(z0, z1)) →
U2_AG(
.(
z0,
z1),
member_out_ag(
z0,
.(
z0,
z1)))
with rule
SUBSET_IN_AG(
.(
x0,
x1)) →
U2_AG(
.(
x0,
x1),
member_out_ag(
x0,
.(
x0,
x1)))
Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence
All these steps are and every following step will be a correct step w.r.t to Q.
(48) FALSE