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SRS Standard pair #516969148
details
property
value
status
complete
benchmark
sym-6.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n169.star.cs.uiowa.edu
space
Waldmann_06_SRS
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
27.3667111397 seconds
cpu usage
105.66921443
max memory
2.52086272E9
stage attributes
key
value
output-size
7650
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 1 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 1308 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 1627 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 937 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 356 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: c(c(c(c(x1)))) -> b(b(b(b(x1)))) b(b(x1)) -> x1 b(b(x1)) -> c(b(c(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: c(c(c(c(x1)))) -> b(b(b(b(x1)))) b(b(x1)) -> x1 b(b(x1)) -> c(b(c(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: C(c(c(c(x1)))) -> B(b(b(b(x1)))) C(c(c(c(x1)))) -> B(b(b(x1))) C(c(c(c(x1)))) -> B(b(x1)) C(c(c(c(x1)))) -> B(x1) B(b(x1)) -> C(b(c(x1))) B(b(x1)) -> B(c(x1)) B(b(x1)) -> C(x1) The TRS R consists of the following rules: c(c(c(c(x1)))) -> b(b(b(b(x1)))) b(b(x1)) -> x1 b(b(x1)) -> c(b(c(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. B(b(x1)) -> C(b(c(x1))) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic integers [ARCTIC,STERNAGEL_THIEMANN_RTA14]: <<<
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