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TRS Standard pair #487074038
details
property
value
status
complete
benchmark
list-sum-prod.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n177.star.cs.uiowa.edu
space
CiME_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.90838 seconds
cpu usage
3.73079
user time
3.57674
system time
0.154052
max virtual memory
1.8408912E7
max residence set size
239756.0
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 102 ms] (2) QTRS (3) RisEmptyProof [EQUIVALENT, 0 ms] (4) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: +(x, 0) -> x +(0, x) -> x +(s(x), s(y)) -> s(s(+(x, y))) *(x, 0) -> 0 *(0, x) -> 0 *(s(x), s(y)) -> s(+(*(x, y), +(x, y))) sum(nil) -> 0 sum(cons(x, l)) -> +(x, sum(l)) prod(nil) -> s(0) prod(cons(x, l)) -> *(x, prod(l)) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: +/2(YES,YES) 0/0) s/1(YES) */2(YES,YES) sum/1(YES) nil/0) cons/2(YES,YES) prod/1)YES( Quasi precedence: [sum_1, cons_2] > [0, *_2] > +_2 > s_1 nil > [0, *_2] > +_2 > s_1 Status: +_2: multiset status 0: multiset status s_1: multiset status *_2: [2,1] sum_1: multiset status nil: multiset status cons_2: multiset status With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: +(x, 0) -> x +(0, x) -> x +(s(x), s(y)) -> s(s(+(x, y))) *(x, 0) -> 0 *(0, x) -> 0 *(s(x), s(y)) -> s(+(*(x, y), +(x, y))) sum(nil) -> 0 sum(cons(x, l)) -> +(x, sum(l)) prod(nil) -> s(0) prod(cons(x, l)) -> *(x, prod(l)) ---------------------------------------- (2) Obligation: Q restricted rewrite system: R is empty. Q is empty. ---------------------------------------- (3) RisEmptyProof (EQUIVALENT) The TRS R is empty. Hence, termination is trivially proven. ---------------------------------------- (4) YES
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