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SRS Standard pair #516968380
details
property
value
status
complete
benchmark
z113.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n083.star.cs.uiowa.edu
space
Zantema_04
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
4.52439284325 seconds
cpu usage
14.471328195
max memory
1.24135424E9
stage attributes
key
value
output-size
6768
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 22 ms] (4) QTRS (5) DependencyPairsProof [EQUIVALENT, 1 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 1 ms] (8) AND (9) QDP (10) UsableRulesProof [EQUIVALENT, 0 ms] (11) QDP (12) MRRProof [EQUIVALENT, 9 ms] (13) QDP (14) PisEmptyProof [EQUIVALENT, 0 ms] (15) YES (16) QDP (17) UsableRulesProof [EQUIVALENT, 0 ms] (18) QDP (19) QDPSizeChangeProof [EQUIVALENT, 1 ms] (20) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 1(1(x1)) -> 4(3(x1)) 1(2(x1)) -> 2(1(x1)) 2(2(x1)) -> 1(1(1(x1))) 3(3(x1)) -> 5(6(x1)) 3(4(x1)) -> 1(1(x1)) 4(4(x1)) -> 3(x1) 5(5(x1)) -> 6(2(x1)) 5(6(x1)) -> 1(2(x1)) 6(6(x1)) -> 2(1(x1)) Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 1(1(x1)) -> 3(4(x1)) 2(1(x1)) -> 1(2(x1)) 2(2(x1)) -> 1(1(1(x1))) 3(3(x1)) -> 6(5(x1)) 4(3(x1)) -> 1(1(x1)) 4(4(x1)) -> 3(x1) 5(5(x1)) -> 2(6(x1)) 6(5(x1)) -> 2(1(x1)) 6(6(x1)) -> 1(2(x1)) Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(1(x_1)) = 85 + x_1 POL(2(x_1)) = 128 + x_1 POL(3(x_1)) = 113 + x_1 POL(4(x_1)) = 57 + x_1 POL(5(x_1)) = 118 + x_1 POL(6(x_1)) = 107 + x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 2(2(x1)) -> 1(1(1(x1))) 3(3(x1)) -> 6(5(x1)) 4(4(x1)) -> 3(x1) 5(5(x1)) -> 2(6(x1)) 6(5(x1)) -> 2(1(x1)) 6(6(x1)) -> 1(2(x1))
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