details

property	value
status	complete
benchmark	dup15.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n176.star.cs.uiowa.edu
space	Trafo_06

run statistics

property	value
solver	ttt2-1.19
configuration	ttt2
runtime (wallclock)	1.8279 seconds
cpu usage	11.4942
user time	9.49777
system time	1.99641
max virtual memory	3617168.0
max residence set size	71228.0

stage attributes

key	value
starexec-result	YES

output

5.62/1.82 YES 5.62/1.82 5.62/1.82 Problem: 5.62/1.82 a(a(d(d(x1)))) -> d(d(b(b(x1)))) 5.62/1.82 a(a(x1)) -> b(b(b(b(b(b(x1)))))) 5.62/1.82 b(b(d(d(b(b(x1)))))) -> a(a(c(c(x1)))) 5.62/1.82 c(c(x1)) -> d(d(x1)) 5.62/1.82 5.62/1.82 Proof: 5.62/1.82 Matrix Interpretation Processor: dim=1 5.62/1.82 5.62/1.82 interpretation: 5.62/1.82 [c](x0) = 2x0, 5.62/1.82 5.62/1.82 [b](x0) = x0 + 1, 5.62/1.82 5.62/1.82 [a](x0) = x0 + 5, 5.62/1.82 5.62/1.82 [d](x0) = 2x0 5.62/1.82 orientation: 5.62/1.82 a(a(d(d(x1)))) = 4x1 + 10 >= 4x1 + 8 = d(d(b(b(x1)))) 5.62/1.82 5.62/1.82 a(a(x1)) = x1 + 10 >= x1 + 6 = b(b(b(b(b(b(x1)))))) 5.62/1.82 5.62/1.82 b(b(d(d(b(b(x1)))))) = 4x1 + 10 >= 4x1 + 10 = a(a(c(c(x1)))) 5.62/1.82 5.62/1.82 c(c(x1)) = 4x1 >= 4x1 = d(d(x1)) 5.62/1.82 problem: 5.62/1.82 b(b(d(d(b(b(x1)))))) -> a(a(c(c(x1)))) 5.62/1.82 c(c(x1)) -> d(d(x1)) 5.62/1.82 KBO Processor: 5.62/1.82 weight function: 5.62/1.82 w0 = 1 5.62/1.82 w(c) = w(b) = w(d) = 1 5.62/1.82 w(a) = 0 5.62/1.82 precedence: 5.62/1.82 a > c > b ~ d 5.62/1.82 problem: 5.62/1.82 5.62/1.82 Qed 5.62/1.82 EOF popout

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job log

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actions

all output return to SRS_Standard 2019-03-29 03.29