Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
TRS Standard pair #487068038
details
property
value
status
complete
benchmark
4.14.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n180.star.cs.uiowa.edu
space
SK90
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.7472 seconds
cpu usage
3.95922
user time
3.78503
system time
0.174191
max virtual memory
1.8343148E7
max residence set size
242336.0
stage attributes
key
value
starexec-result
YES
output
YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 60 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 1 ms] (4) QTRS (5) RisEmptyProof [EQUIVALENT, 0 ms] (6) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: p(s(x)) -> x s(p(x)) -> x +(0, y) -> y +(s(x), y) -> s(+(x, y)) +(p(x), y) -> p(+(x, y)) minus(0) -> 0 minus(s(x)) -> p(minus(x)) minus(p(x)) -> s(minus(x)) *(0, y) -> 0 *(s(x), y) -> +(*(x, y), y) *(p(x), y) -> +(*(x, y), minus(y)) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: p/1(YES) s/1(YES) +/2(YES,YES) 0/0) minus/1)YES( */2(YES,YES) Quasi precedence: *_2 > +_2 > [p_1, s_1] > 0 Status: p_1: multiset status s_1: multiset status +_2: multiset status 0: multiset status *_2: multiset status With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: p(s(x)) -> x s(p(x)) -> x +(0, y) -> y +(s(x), y) -> s(+(x, y)) +(p(x), y) -> p(+(x, y)) *(0, y) -> 0 *(s(x), y) -> +(*(x, y), y) *(p(x), y) -> +(*(x, y), minus(y)) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: minus(0) -> 0 minus(s(x)) -> p(minus(x)) minus(p(x)) -> s(minus(x)) Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Knuth-Bendix order [KBO] with precedence:minus_1 > p_1 > 0 > s_1 and weight map: 0=1 minus_1=2 s_1=1 p_1=1 The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to TRS Standard