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TRS Stand 20472 pair #381713132
details
property
value
status
complete
benchmark
revlist.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n091.star.cs.uiowa.edu
space
Rubio_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.93588709831 seconds
cpu usage
4.563194558
max memory
2.25325056E8
stage attributes
key
value
output-size
8336
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) QDPOrderProof [EQUIVALENT, 39 ms] (16) QDP (17) DependencyGraphProof [EQUIVALENT, 0 ms] (18) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: rev1(0, nil) -> 0 rev1(s(X), nil) -> s(X) rev1(X, cons(Y, L)) -> rev1(Y, L) rev(nil) -> nil rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L)) rev2(X, nil) -> nil rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L)))) Q is empty. ---------------------------------------- (1) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: rev1(0, nil) -> 0 rev1(s(X), nil) -> s(X) rev1(X, cons(Y, L)) -> rev1(Y, L) rev(nil) -> nil rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L)) rev2(X, nil) -> nil rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L)))) The set Q consists of the following terms: rev1(0, nil) rev1(s(x0), nil) rev1(x0, cons(x1, x2)) rev(nil) rev(cons(x0, x1)) rev2(x0, nil) rev2(x0, cons(x1, x2)) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: REV1(X, cons(Y, L)) -> REV1(Y, L) REV(cons(X, L)) -> REV1(X, L) REV(cons(X, L)) -> REV2(X, L) REV2(X, cons(Y, L)) -> REV(cons(X, rev(rev2(Y, L)))) REV2(X, cons(Y, L)) -> REV(rev2(Y, L)) REV2(X, cons(Y, L)) -> REV2(Y, L) The TRS R consists of the following rules:
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