Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
SRS Standard pair #516973342
details
property
value
status
complete
benchmark
14.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n076.star.cs.uiowa.edu
space
Gebhardt_06
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
15.4798970222 seconds
cpu usage
47.652164069
max memory
1.806364672E9
stage attributes
key
value
output-size
3906
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 10 ms] (2) QDP (3) MRRProof [EQUIVALENT, 80 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 323 ms] (6) QDP (7) PisEmptyProof [EQUIVALENT, 0 ms] (8) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(0(0(0(x1)))) -> 0(1(1(1(x1)))) 1(1(0(1(x1)))) -> 0(0(0(0(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: 0^1(0(0(0(x1)))) -> 0^1(1(1(1(x1)))) 0^1(0(0(0(x1)))) -> 1^1(1(1(x1))) 0^1(0(0(0(x1)))) -> 1^1(1(x1)) 0^1(0(0(0(x1)))) -> 1^1(x1) 1^1(1(0(1(x1)))) -> 0^1(0(0(0(x1)))) 1^1(1(0(1(x1)))) -> 0^1(0(0(x1))) 1^1(1(0(1(x1)))) -> 0^1(0(x1)) 1^1(1(0(1(x1)))) -> 0^1(x1) The TRS R consists of the following rules: 0(0(0(0(x1)))) -> 0(1(1(1(x1)))) 1(1(0(1(x1)))) -> 0(0(0(0(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented dependency pairs: 0^1(0(0(0(x1)))) -> 1^1(1(1(x1))) 0^1(0(0(0(x1)))) -> 1^1(1(x1)) 0^1(0(0(0(x1)))) -> 1^1(x1) 1^1(1(0(1(x1)))) -> 0^1(0(0(0(x1)))) 1^1(1(0(1(x1)))) -> 0^1(0(0(x1))) 1^1(1(0(1(x1)))) -> 0^1(0(x1)) 1^1(1(0(1(x1)))) -> 0^1(x1) Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = 2 + x_1 POL(0^1(x_1)) = 1 + x_1 POL(1(x_1)) = 2 + x_1 POL(1^1(x_1)) = 2 + x_1 ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: 0^1(0(0(0(x1)))) -> 0^1(1(1(1(x1)))) The TRS R consists of the following rules: 0(0(0(0(x1)))) -> 0(1(1(1(x1)))) 1(1(0(1(x1)))) -> 0(0(0(0(x1))))
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to SRS Standard