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SRS Stand 10685 pair #381720280
details
property
value
status
complete
benchmark
z100.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n027.star.cs.uiowa.edu
space
Zantema_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
7.57855796814 seconds
cpu usage
26.1504181
max memory
3.829628928E9
stage attributes
key
value
output-size
12605
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 31 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 2 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) MRRProof [EQUIVALENT, 25 ms] (9) QDP (10) PisEmptyProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) QDPOrderProof [EQUIVALENT, 0 ms] (16) QDP (17) PisEmptyProof [EQUIVALENT, 0 ms] (18) YES (19) QDP (20) UsableRulesProof [EQUIVALENT, 0 ms] (21) QDP (22) QDPSizeChangeProof [EQUIVALENT, 0 ms] (23) YES (24) QDP (25) QDPOrderProof [EQUIVALENT, 308 ms] (26) QDP (27) QDPOrderProof [EQUIVALENT, 0 ms] (28) QDP (29) PisEmptyProof [EQUIVALENT, 0 ms] (30) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: r1(a(x1)) -> a(a(a(r1(x1)))) r2(a(x1)) -> a(a(a(r2(x1)))) a(l1(x1)) -> l1(a(a(a(x1)))) a(a(l2(x1))) -> l2(a(a(x1))) r1(b(x1)) -> l1(b(x1)) r2(b(x1)) -> l2(a(b(x1))) b(l1(x1)) -> b(r2(x1)) b(l2(x1)) -> b(r1(x1)) a(a(x1)) -> x1 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: R1(a(x1)) -> A(a(a(r1(x1)))) R1(a(x1)) -> A(a(r1(x1))) R1(a(x1)) -> A(r1(x1)) R1(a(x1)) -> R1(x1) R2(a(x1)) -> A(a(a(r2(x1)))) R2(a(x1)) -> A(a(r2(x1))) R2(a(x1)) -> A(r2(x1)) R2(a(x1)) -> R2(x1) A(l1(x1)) -> A(a(a(x1))) A(l1(x1)) -> A(a(x1)) A(l1(x1)) -> A(x1) A(a(l2(x1))) -> A(a(x1)) A(a(l2(x1))) -> A(x1) R2(b(x1)) -> A(b(x1)) B(l1(x1)) -> B(r2(x1)) B(l1(x1)) -> R2(x1) B(l2(x1)) -> B(r1(x1)) B(l2(x1)) -> R1(x1) The TRS R consists of the following rules: r1(a(x1)) -> a(a(a(r1(x1)))) r2(a(x1)) -> a(a(a(r2(x1)))) a(l1(x1)) -> l1(a(a(a(x1))))
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