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TRS Innermost pair #487093387
details
property
value
status
complete
benchmark
Ex2_Luc02a_L.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n140.star.cs.uiowa.edu
space
Transformed_CSR_innermost_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.75239 seconds
cpu usage
3.78383
user time
3.61601
system time
0.167824
max virtual memory
1.8343376E7
max residence set size
239976.0
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 56 ms] (2) QTRS (3) RisEmptyProof [EQUIVALENT, 0 ms] (4)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) The set Q consists of the following terms: terms(x0) sqr(0) sqr(s(x0)) dbl(0) dbl(s(x0)) add(0, x0) add(s(x0), x1) first(0, x0) first(s(x0), cons(x1)) ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Quasi precedence: terms_1 > [cons_1, first_2] terms_1 > recip_1 terms_1 > sqr_1 > 0 > nil terms_1 > sqr_1 > add_2 > s_1 terms_1 > sqr_1 > dbl_1 > s_1 Status: terms_1: [1] cons_1: [1] recip_1: multiset status sqr_1: multiset status 0: multiset status s_1: multiset status add_2: multiset status dbl_1: multiset status first_2: multiset status nil: 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) ---------------------------------------- (2) Obligation: Q restricted rewrite system: R is empty. The set Q consists of the following terms: terms(x0) sqr(0) sqr(s(x0)) dbl(0) dbl(s(x0)) add(0, x0) add(s(x0), x1) first(0, x0) first(s(x0), cons(x1))
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