Logic Programming pair #487096152

details

property	value
status	complete
benchmark	subset.pl
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n137.star.cs.uiowa.edu
space	talp_apt

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.00883 seconds
cpu usage	4.41685
user time	4.18923
system time	0.227622
max virtual memory	1.900996E7
max residence set size	371060.0

stage attributes

key	value
starexec-result	YES

output

YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.pl
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


Left Termination of the query pattern

subset(g,g)

w.r.t. the given Prolog program could successfully be proven:

(0) Prolog
(1) PrologToPiTRSProof [SOUND, 0 ms]
(2) PiTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) PiDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) PiDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) PiDP
        (10) PiDPToQDPProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 1 ms]
        (13) YES
    (14) PiDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) PiDP
        (17) PiDPToQDPProof [SOUND, 0 ms]
        (18) QDP
        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (20) YES


----------------------------------------

(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)
We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:

subset_in_2: (b,b)

member_in_2: (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

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