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TRS Standard pair #487070143
details
property
value
status
complete
benchmark
ExSec11_1_Luc02a_L.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n173.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.85307 seconds
cpu usage
3.70301
user time
3.53932
system time
0.163693
max virtual memory
1.8408912E7
max residence set size
235856.0
stage attributes
key
value
starexec-result
YES
output
YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 42 ms] (2) QTRS (3) RisEmptyProof [EQUIVALENT, 0 ms] (4) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: terms(N) -> cons(recip(sqr(N))) sqr(0) -> 0 sqr(s(X)) -> s(add(sqr(X), dbl(X))) dbl(0) -> 0 dbl(s(X)) -> s(s(dbl(X))) add(0, X) -> X add(s(X), Y) -> s(add(X, Y)) first(0, X) -> nil first(s(X), cons(Y)) -> cons(Y) half(0) -> 0 half(s(0)) -> 0 half(s(s(X))) -> s(half(X)) half(dbl(X)) -> X Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Quasi precedence: terms_1 > cons_1 > s_1 terms_1 > recip_1 > s_1 terms_1 > [sqr_1, dbl_1] > [0, nil, half_1] > s_1 terms_1 > [sqr_1, dbl_1] > add_2 > s_1 first_2 > s_1 Status: terms_1: multiset status cons_1: multiset status recip_1: multiset status sqr_1: multiset status 0: multiset status s_1: multiset status add_2: [1,2] dbl_1: multiset status first_2: multiset status nil: multiset status half_1: multiset status With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: terms(N) -> cons(recip(sqr(N))) sqr(0) -> 0 sqr(s(X)) -> s(add(sqr(X), dbl(X))) dbl(0) -> 0 dbl(s(X)) -> s(s(dbl(X))) add(0, X) -> X add(s(X), Y) -> s(add(X, Y)) first(0, X) -> nil first(s(X), cons(Y)) -> cons(Y) half(0) -> 0 half(s(0)) -> 0 half(s(s(X))) -> s(half(X)) half(dbl(X)) -> X ---------------------------------------- (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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