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TRS Stand 20472 pair #381710136
details
property
value
status
complete
benchmark
intersect.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n104.star.cs.uiowa.edu
space
CiME_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.77358603477 seconds
cpu usage
3.88928321
max memory
2.5038848E8
stage attributes
key
value
output-size
13097
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 2 ms] (7) QDP (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) UsableRulesProof [EQUIVALENT, 0 ms] (12) QDP (13) QDPSizeChangeProof [EQUIVALENT, 0 ms] (14) YES (15) QDP (16) QDPSizeChangeProof [EQUIVALENT, 0 ms] (17) YES (18) QDP (19) QDPSizeChangeProof [EQUIVALENT, 1 ms] (20) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: if(true, x, y) -> x if(false, x, y) -> y eq(0, 0) -> true eq(0, s(x)) -> false eq(s(x), 0) -> false eq(s(x), s(y)) -> eq(x, y) app(nil, l) -> l app(cons(x, l1), l2) -> cons(x, app(l1, l2)) app(app(l1, l2), l3) -> app(l1, app(l2, l3)) mem(x, nil) -> false mem(x, cons(y, l)) -> ifmem(eq(x, y), x, l) ifmem(true, x, l) -> true ifmem(false, x, l) -> mem(x, l) inter(x, nil) -> nil inter(nil, x) -> nil inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3)) inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3)) inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2) inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1) ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2)) ifinter(false, x, l1, l2) -> inter(l1, l2) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: EQ(s(x), s(y)) -> EQ(x, y) APP(cons(x, l1), l2) -> APP(l1, l2) APP(app(l1, l2), l3) -> APP(l1, app(l2, l3)) APP(app(l1, l2), l3) -> APP(l2, l3) MEM(x, cons(y, l)) -> IFMEM(eq(x, y), x, l) MEM(x, cons(y, l)) -> EQ(x, y) IFMEM(false, x, l) -> MEM(x, l) INTER(app(l1, l2), l3) -> APP(inter(l1, l3), inter(l2, l3)) INTER(app(l1, l2), l3) -> INTER(l1, l3) INTER(app(l1, l2), l3) -> INTER(l2, l3) INTER(l1, app(l2, l3)) -> APP(inter(l1, l2), inter(l1, l3)) INTER(l1, app(l2, l3)) -> INTER(l1, l2) INTER(l1, app(l2, l3)) -> INTER(l1, l3) INTER(cons(x, l1), l2) -> IFINTER(mem(x, l2), x, l1, l2) INTER(cons(x, l1), l2) -> MEM(x, l2) INTER(l1, cons(x, l2)) -> IFINTER(mem(x, l1), x, l2, l1) INTER(l1, cons(x, l2)) -> MEM(x, l1) IFINTER(true, x, l1, l2) -> INTER(l1, l2) IFINTER(false, x, l1, l2) -> INTER(l1, l2) The TRS R consists of the following rules:
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