details

property	value
status	complete
benchmark	rt1-5.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n049.star.cs.uiowa.edu
space	Relative_05

run statistics

property	value
solver	ttt2-1.17+nonreach
configuration	ttt2-1.17+nonreach
runtime (wallclock)	2.36567902565 seconds
cpu usage	8.229356571
max memory	8.02537472E8

stage attributes

key	value
output-size	3510
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: strict: s(a(x)) -> s(b(x)) b(b(x)) -> a(x) weak: f(s(x),y) -> f(x,s(y)) s(a(x)) -> a(s(x)) s(b(x)) -> b(s(x)) a(s(x)) -> s(a(x)) b(s(x)) -> s(b(x)) Proof: Matrix Interpretation Processor: dim=2 interpretation: [2 0] [1 1] [0] [f](x0, x1) = [0 0]x0 + [3 0]x1 + [2], [1 0] [1] [b](x0) = [0 0]x0 + [0], [1 0] [s](x0) = [0 0]x0, [1 0] [1] [a](x0) = [0 0]x0 + [0] orientation: [1 0] [1] [1 0] [1] s(a(x)) = [0 0]x + [0] >= [0 0]x + [0] = s(b(x)) [1 0] [2] [1 0] [1] b(b(x)) = [0 0]x + [0] >= [0 0]x + [0] = a(x) [2 0] [1 1] [0] [2 0] [1 0] [0] f(s(x),y) = [0 0]x + [3 0]y + [2] >= [0 0]x + [3 0]y + [2] = f(x,s(y)) [1 0] [1] [1 0] [1] s(a(x)) = [0 0]x + [0] >= [0 0]x + [0] = a(s(x)) [1 0] [1] [1 0] [1] s(b(x)) = [0 0]x + [0] >= [0 0]x + [0] = b(s(x)) [1 0] [1] [1 0] [1] a(s(x)) = [0 0]x + [0] >= [0 0]x + [0] = s(a(x)) [1 0] [1] [1 0] [1] b(s(x)) = [0 0]x + [0] >= [0 0]x + [0] = s(b(x)) problem: strict: s(a(x)) -> s(b(x)) weak: f(s(x),y) -> f(x,s(y)) s(a(x)) -> a(s(x)) s(b(x)) -> b(s(x)) a(s(x)) -> s(a(x)) b(s(x)) -> s(b(x)) Matrix Interpretation Processor: dim=2 interpretation: [2 2] [2 1] [1] [f](x0, x1) = [0 0]x0 + [0 0]x1 + [2], [b](x0) = x0, [0] [s](x0) = x0 + [1], [0] [a](x0) = x0 + [2] orientation: [0] [0] s(a(x)) = x + [3] >= x + [1] = s(b(x)) [2 2] [2 1] [3] [2 2] [2 1] [2] f(s(x),y) = [0 0]x + [0 0]y + [2] >= [0 0]x + [0 0]y + [2] = f(x,s(y)) [0] [0] s(a(x)) = x + [3] >= x + [3] = a(s(x)) [0] [0] s(b(x)) = x + [1] >= x + [1] = b(s(x)) [0] [0] a(s(x)) = x + [3] >= x + [3] = s(a(x)) [0] [0] b(s(x)) = x + [1] >= x + [1] = s(b(x)) problem: strict: s(a(x)) -> s(b(x)) weak: popout

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

job log

popout

actions

all output return to TRS Relat 75837