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Logic Progr 19030 pair #381919989
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
n057.star.cs.uiowa.edu
space
talp_apt
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.73691987991 seconds
cpu usage
4.021287689
max memory
2.81382912E8
stage attributes
key
value
output-size
13643
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.pl /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.pl # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern subset(g,g) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) PrologToPiTRSProof [SOUND,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 [SOUND, 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). 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
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