Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
TRS Standard pair #516964505
details
property
value
status
complete
benchmark
perfect.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n054.star.cs.uiowa.edu
space
Mixed_TRS
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
2.31186389923 seconds
cpu usage
6.134350735
max memory
4.8037888E8
stage attributes
key
value
output-size
5668
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 9 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) QReductionProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: perfectp(0) -> false perfectp(s(x)) -> f(x, s(0), s(x), s(x)) f(0, y, 0, u) -> true f(0, y, s(z), u) -> false f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u) f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) Q is empty. ---------------------------------------- (1) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: perfectp(0) -> false perfectp(s(x)) -> f(x, s(0), s(x), s(x)) f(0, y, 0, u) -> true f(0, y, s(z), u) -> false f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u) f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) The set Q consists of the following terms: perfectp(0) perfectp(s(x0)) f(0, x0, 0, x1) f(0, x0, s(x1), x2) f(s(x0), 0, x1, x2) f(s(x0), s(x1), x2, x3) ---------------------------------------- (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: PERFECTP(s(x)) -> F(x, s(0), s(x), s(x)) F(s(x), 0, z, u) -> F(x, u, minus(z, s(x)), u) F(s(x), s(y), z, u) -> F(s(x), minus(y, x), z, u) F(s(x), s(y), z, u) -> F(x, u, z, u) The TRS R consists of the following rules: perfectp(0) -> false perfectp(s(x)) -> f(x, s(0), s(x), s(x)) f(0, y, 0, u) -> true f(0, y, s(z), u) -> false f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u) f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) The set Q consists of the following terms: perfectp(0)
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to TRS Standard