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SRS Stand 10685 pair #381718087
details
property
value
status
complete
benchmark
11.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n039.star.cs.uiowa.edu
space
Bouchare_06
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
8.99105000496 seconds
cpu usage
31.342905366
max memory
3.4049024E9
stage attributes
key
value
output-size
7575
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 6 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 102 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 50 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 26 ms] (10) QDP (11) DependencyGraphProof [EQUIVALENT, 0 ms] (12) QDP (13) QDPOrderProof [EQUIVALENT, 293 ms] (14) QDP (15) PisEmptyProof [EQUIVALENT, 0 ms] (16) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(a(x1))) -> b(x1) b(b(x1)) -> a(a(x1)) a(a(x1)) -> a(b(a(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: a(a(a(x1))) -> b(x1) b(b(x1)) -> a(a(x1)) a(a(x1)) -> a(b(a(x1))) Q is empty. ---------------------------------------- (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: A(a(a(x1))) -> B(x1) B(b(x1)) -> A(a(x1)) B(b(x1)) -> A(x1) A(a(x1)) -> A(b(a(x1))) A(a(x1)) -> B(a(x1)) The TRS R consists of the following rules: a(a(a(x1))) -> b(x1) b(b(x1)) -> a(a(x1)) a(a(x1)) -> a(b(a(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. A(a(a(x1))) -> B(x1) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<<
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