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TRS Standard pair #516962380
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
n005.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
2.14306783676 seconds
cpu usage
5.721014856
max memory
4.7521792E8
stage attributes
key
value
output-size
2291
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 91 ms] (2) QTRS (3) RisEmptyProof [EQUIVALENT, 0 ms] (4) YES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. ----------------------------------------
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