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TRS Stand 20472 pair #381710494
details
property
value
status
complete
benchmark
ExIntrod_GM04_GM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n085.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.09979176521 seconds
cpu usage
5.288902204
max memory
3.43515136E8
stage attributes
key
value
output-size
11287
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 98 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 0 ms] (4) QTRS (5) QTRSRRRProof [EQUIVALENT, 14 ms] (6) QTRS (7) QTRSRRRProof [EQUIVALENT, 0 ms] (8) QTRS (9) QTRSRRRProof [EQUIVALENT, 9 ms] (10) QTRS (11) QTRSRRRProof [EQUIVALENT, 0 ms] (12) QTRS (13) QTRSRRRProof [EQUIVALENT, 8 ms] (14) QTRS (15) QTRSRRRProof [EQUIVALENT, 0 ms] (16) QTRS (17) QTRSRRRProof [EQUIVALENT, 5 ms] (18) QTRS (19) RisEmptyProof [EQUIVALENT, 0 ms] (20) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__nats -> a__adx(a__zeros) a__zeros -> cons(0, zeros) a__incr(cons(X, Y)) -> cons(s(X), incr(Y)) a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y))) a__hd(cons(X, Y)) -> mark(X) a__tl(cons(X, Y)) -> mark(Y) mark(nats) -> a__nats mark(adx(X)) -> a__adx(mark(X)) mark(zeros) -> a__zeros mark(incr(X)) -> a__incr(mark(X)) mark(hd(X)) -> a__hd(mark(X)) mark(tl(X)) -> a__tl(mark(X)) mark(cons(X1, X2)) -> cons(X1, X2) mark(0) -> 0 mark(s(X)) -> s(X) a__nats -> nats a__adx(X) -> adx(X) a__zeros -> zeros a__incr(X) -> incr(X) a__hd(X) -> hd(X) a__tl(X) -> tl(X) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(a__adx(x_1)) = x_1 POL(a__hd(x_1)) = 1 + 2*x_1 POL(a__incr(x_1)) = x_1 POL(a__nats) = 0 POL(a__tl(x_1)) = 1 + 2*x_1 POL(a__zeros) = 0 POL(adx(x_1)) = x_1 POL(cons(x_1, x_2)) = x_1 + 2*x_2 POL(hd(x_1)) = 1 + 2*x_1 POL(incr(x_1)) = x_1 POL(mark(x_1)) = 2*x_1 POL(nats) = 0 POL(s(x_1)) = x_1 POL(tl(x_1)) = 1 + 2*x_1 POL(zeros) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: a__hd(cons(X, Y)) -> mark(X) a__tl(cons(X, Y)) -> mark(Y) mark(hd(X)) -> a__hd(mark(X)) mark(tl(X)) -> a__tl(mark(X)) ----------------------------------------
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