details

property	value
status	complete
benchmark	z075.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n161.star.cs.uiowa.edu
space	Zantema_04

run statistics

property	value
solver	ttt2-1.19
configuration	ttt2
runtime (wallclock)	3.01345 seconds
cpu usage	10.5396
user time	8.0653
system time	2.47426
max virtual memory	3577176.0
max residence set size	68944.0

stage attributes

key	value
starexec-result	YES

output

10.27/2.99 YES 10.27/3.00 10.27/3.00 Problem: 10.27/3.00 t(f(x1)) -> t(c(n(x1))) 10.27/3.00 n(f(x1)) -> f(n(x1)) 10.27/3.00 o(f(x1)) -> f(o(x1)) 10.27/3.00 n(s(x1)) -> f(s(x1)) 10.27/3.00 o(s(x1)) -> f(s(x1)) 10.27/3.00 c(f(x1)) -> f(c(x1)) 10.27/3.00 c(n(x1)) -> n(c(x1)) 10.27/3.00 c(o(x1)) -> o(c(x1)) 10.27/3.00 c(o(x1)) -> o(x1) 10.27/3.00 10.27/3.00 Proof: 10.27/3.00 String Reversal Processor: 10.27/3.00 f(t(x1)) -> n(c(t(x1))) 10.27/3.00 f(n(x1)) -> n(f(x1)) 10.27/3.00 f(o(x1)) -> o(f(x1)) 10.27/3.00 s(n(x1)) -> s(f(x1)) 10.27/3.00 s(o(x1)) -> s(f(x1)) 10.27/3.00 f(c(x1)) -> c(f(x1)) 10.27/3.00 n(c(x1)) -> c(n(x1)) 10.27/3.00 o(c(x1)) -> c(o(x1)) 10.27/3.00 o(c(x1)) -> o(x1) 10.27/3.00 Matrix Interpretation Processor: dim=1 10.27/3.00 10.27/3.00 interpretation: 10.27/3.00 [s](x0) = x0 + 8, 10.27/3.00 10.27/3.00 [o](x0) = 4x0 + 8, 10.27/3.00 10.27/3.00 [c](x0) = x0, 10.27/3.00 10.27/3.00 [n](x0) = 2x0 + 2, 10.27/3.00 10.27/3.00 [t](x0) = 8x0, 10.27/3.00 10.27/3.00 [f](x0) = 2x0 + 2 10.27/3.00 orientation: 10.27/3.00 f(t(x1)) = 16x1 + 2 >= 16x1 + 2 = n(c(t(x1))) 10.27/3.00 10.27/3.00 f(n(x1)) = 4x1 + 6 >= 4x1 + 6 = n(f(x1)) 10.27/3.00 10.27/3.00 f(o(x1)) = 8x1 + 18 >= 8x1 + 16 = o(f(x1)) 10.27/3.00 10.27/3.00 s(n(x1)) = 2x1 + 10 >= 2x1 + 10 = s(f(x1)) 10.27/3.00 10.27/3.00 s(o(x1)) = 4x1 + 16 >= 2x1 + 10 = s(f(x1)) 10.27/3.00 10.27/3.00 f(c(x1)) = 2x1 + 2 >= 2x1 + 2 = c(f(x1)) 10.27/3.00 10.27/3.00 n(c(x1)) = 2x1 + 2 >= 2x1 + 2 = c(n(x1)) 10.27/3.00 10.27/3.00 o(c(x1)) = 4x1 + 8 >= 4x1 + 8 = c(o(x1)) 10.27/3.00 10.27/3.00 o(c(x1)) = 4x1 + 8 >= 4x1 + 8 = o(x1) 10.27/3.00 problem: 10.27/3.00 f(t(x1)) -> n(c(t(x1))) 10.27/3.00 f(n(x1)) -> n(f(x1)) 10.27/3.00 s(n(x1)) -> s(f(x1)) 10.27/3.00 f(c(x1)) -> c(f(x1)) 10.27/3.00 n(c(x1)) -> c(n(x1)) 10.27/3.00 o(c(x1)) -> c(o(x1)) 10.27/3.00 o(c(x1)) -> o(x1) 10.27/3.00 String Reversal Processor: 10.27/3.00 t(f(x1)) -> t(c(n(x1))) 10.27/3.00 n(f(x1)) -> f(n(x1)) 10.27/3.00 n(s(x1)) -> f(s(x1)) 10.27/3.00 c(f(x1)) -> f(c(x1)) 10.27/3.00 c(n(x1)) -> n(c(x1)) 10.27/3.00 c(o(x1)) -> o(c(x1)) 10.27/3.00 c(o(x1)) -> o(x1) 10.27/3.00 Matrix Interpretation Processor: dim=3 10.27/3.00 10.27/3.00 interpretation: 10.27/3.00 [1 0 0] [0] 10.27/3.00 [s](x0) = [0 0 0]x0 + [0] 10.27/3.00 [0 0 0] [1], 10.27/3.00 10.27/3.00 [1 1 0] 10.27/3.00 [o](x0) = [0 0 0]x0 10.27/3.00 [0 0 0] , 10.27/3.00 10.27/3.00 [1 1 0] 10.27/3.00 [c](x0) = [0 0 0]x0 10.27/3.00 [0 0 0] , 10.27/3.00 10.27/3.00 [1 0 1] 10.27/3.00 [n](x0) = [0 1 0]x0 10.27/3.00 [0 0 1] , 10.27/3.00 10.27/3.00 [1 0 1] [0] 10.27/3.00 [t](x0) = [0 0 0]x0 + [0] 10.27/3.00 [0 0 1] [1], 10.27/3.00 10.27/3.00 [1 1 0] 10.27/3.00 [f](x0) = [0 0 0]x0 10.27/3.00 [0 0 1] 10.27/3.00 orientation: 10.27/3.00 [1 1 1] [0] [1 1 1] [0] popout

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

job log

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actions

all output return to SRS_Standard 2019-03-29 03.29