Logic Programming pair #487096384

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	n138.star.cs.uiowa.edu
space	terminweb_old

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.79417 seconds
cpu usage	3.89369
user time	3.69714
system time	0.196555
max virtual memory	1.8544588E7
max residence set size	298816.0

stage attributes

key	value
starexec-result	YES

output

YES
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 0 ms]
        (13) YES
    (14) PiDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) PiDP
        (17) PiDPToQDPProof [EQUIVALENT, 1 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).


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)

Pi is empty.



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)

Pi is empty.

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

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

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