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SRS Standard pair #516973096
details
property
value
status
complete
benchmark
size-11-alpha-3-num-14.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n064.star.cs.uiowa.edu
space
Waldmann_07_size11
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
5.12269592285 seconds
cpu usage
16.599271587
max memory
1.290346496E9
stage attributes
key
value
output-size
4555
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 1 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 103 ms] (8) QDP (9) UsableRulesProof [EQUIVALENT, 1 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(x1)) -> x1 a(b(x1)) -> c(c(x1)) c(b(x1)) -> b(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(x1)) -> x1 b(a(x1)) -> c(c(x1)) b(c(x1)) -> a(b(b(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: B(c(x1)) -> A(b(b(x1))) B(c(x1)) -> B(b(x1)) B(c(x1)) -> B(x1) The TRS R consists of the following rules: a(a(x1)) -> x1 b(a(x1)) -> c(c(x1)) b(c(x1)) -> a(b(b(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: B(c(x1)) -> B(x1) B(c(x1)) -> B(b(x1)) The TRS R consists of the following rules: a(a(x1)) -> x1 b(a(x1)) -> c(c(x1)) b(c(x1)) -> a(b(b(x1)))
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