details

property	value
status	complete
benchmark	z009.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n104.star.cs.uiowa.edu
space	Zantema_04

run statistics

property	value
solver	ttt2-1.17+nonreach
configuration	ttt2-1.17+nonreach
runtime (wallclock)	1.35843491554 seconds
cpu usage	4.182823679
max memory	4.13007872E8

stage attributes

key	value
output-size	3106
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES Problem: a(x1) -> b(c(x1)) a(b(x1)) -> b(a(x1)) d(c(x1)) -> d(a(x1)) a(c(x1)) -> c(a(x1)) Proof: Matrix Interpretation Processor: dim=3 interpretation: [1 1 0] [d](x0) = [0 1 0]x0 [0 0 0] , [1 0 0] [b](x0) = [0 0 0]x0 [0 0 1] , [1 0 1] [0] [c](x0) = [0 1 1]x0 + [1] [0 0 0] [0], [1 0 1] [a](x0) = [0 1 1]x0 [0 0 0] orientation: [1 0 1] [1 0 1] a(x1) = [0 1 1]x1 >= [0 0 0]x1 = b(c(x1)) [0 0 0] [0 0 0] [1 0 1] [1 0 1] a(b(x1)) = [0 0 1]x1 >= [0 0 0]x1 = b(a(x1)) [0 0 0] [0 0 0] [1 1 2] [1] [1 1 2] d(c(x1)) = [0 1 1]x1 + [1] >= [0 1 1]x1 = d(a(x1)) [0 0 0] [0] [0 0 0] [1 0 1] [0] [1 0 1] [0] a(c(x1)) = [0 1 1]x1 + [1] >= [0 1 1]x1 + [1] = c(a(x1)) [0 0 0] [0] [0 0 0] [0] problem: a(x1) -> b(c(x1)) a(b(x1)) -> b(a(x1)) a(c(x1)) -> c(a(x1)) String Reversal Processor: a(x1) -> c(b(x1)) b(a(x1)) -> a(b(x1)) c(a(x1)) -> a(c(x1)) Matrix Interpretation Processor: dim=4 interpretation: [1 0 0 0] [0] [1 0 0 1] [1] [b](x0) = [1 0 0 0]x0 + [1] [0 0 0 0] [0], [1 0 0 0] [0 0 0 0] [c](x0) = [0 0 0 0]x0 [0 0 0 0] , [1 0 0 0] [1] [0 0 0 0] [0] [a](x0) = [0 0 1 0]x0 + [0] [0 0 0 1] [0] orientation: [1 0 0 0] [1] [1 0 0 0] [0 0 0 0] [0] [0 0 0 0] a(x1) = [0 0 1 0]x1 + [0] >= [0 0 0 0]x1 = c(b(x1)) [0 0 0 1] [0] [0 0 0 0] [1 0 0 0] [1] [1 0 0 0] [1] [1 0 0 1] [2] [0 0 0 0] [0] b(a(x1)) = [1 0 0 0]x1 + [2] >= [1 0 0 0]x1 + [1] = a(b(x1)) [0 0 0 0] [0] [0 0 0 0] [0] [1 0 0 0] [1] [1 0 0 0] [1] [0 0 0 0] [0] [0 0 0 0] [0] c(a(x1)) = [0 0 0 0]x1 + [0] >= [0 0 0 0]x1 + [0] = a(c(x1)) [0 0 0 0] [0] [0 0 0 0] [0] problem: b(a(x1)) -> a(b(x1)) c(a(x1)) -> a(c(x1)) String Reversal Processor: a(b(x1)) -> b(a(x1)) a(c(x1)) -> c(a(x1)) KBO Processor: weight function: w0 = 1 w(b) = w(c) = 1 w(a) = 0 popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to SRS Stand 10685