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SRS_Standard 2019-03-29 03.29 pair #432294336
details
property
value
status
complete
benchmark
hom02.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n092.star.cs.uiowa.edu
space
Trafo_06
run statistics
property
value
solver
ttt2-1.19
configuration
ttt2
runtime (wallclock)
1.97851 seconds
cpu usage
6.42871
user time
4.99099
system time
1.43772
max virtual memory
3546176.0
max residence set size
69628.0
stage attributes
key
value
starexec-result
YES
output
5.95/1.97 YES 5.95/1.97 5.95/1.97 Problem: 5.95/1.97 a(x1) -> b(b(x1)) 5.95/1.97 a(b(b(x1))) -> b(b(c(c(c(a(x1)))))) 5.95/1.97 b(b(x1)) -> c(c(c(x1))) 5.95/1.97 c(c(c(b(b(x1))))) -> a(x1) 5.95/1.97 5.95/1.97 Proof: 5.95/1.97 String Reversal Processor: 5.95/1.97 a(x1) -> b(b(x1)) 5.95/1.97 b(b(a(x1))) -> a(c(c(c(b(b(x1)))))) 5.95/1.97 b(b(x1)) -> c(c(c(x1))) 5.95/1.97 b(b(c(c(c(x1))))) -> a(x1) 5.95/1.97 Matrix Interpretation Processor: dim=1 5.95/1.97 5.95/1.97 interpretation: 5.95/1.97 [c](x0) = x0, 5.95/1.97 5.95/1.97 [b](x0) = 2x0 + 1, 5.95/1.97 5.95/1.97 [a](x0) = 4x0 + 3 5.95/1.97 orientation: 5.95/1.97 a(x1) = 4x1 + 3 >= 4x1 + 3 = b(b(x1)) 5.95/1.97 5.95/1.97 b(b(a(x1))) = 16x1 + 15 >= 16x1 + 15 = a(c(c(c(b(b(x1)))))) 5.95/1.97 5.95/1.97 b(b(x1)) = 4x1 + 3 >= x1 = c(c(c(x1))) 5.95/1.97 5.95/1.97 b(b(c(c(c(x1))))) = 4x1 + 3 >= 4x1 + 3 = a(x1) 5.95/1.97 problem: 5.95/1.97 a(x1) -> b(b(x1)) 5.95/1.97 b(b(a(x1))) -> a(c(c(c(b(b(x1)))))) 5.95/1.97 b(b(c(c(c(x1))))) -> a(x1) 5.95/1.97 Matrix Interpretation Processor: dim=1 5.95/1.97 5.95/1.97 interpretation: 5.95/1.97 [c](x0) = x0 + 1, 5.95/1.97 5.95/1.97 [b](x0) = 2x0, 5.95/1.97 5.95/1.97 [a](x0) = 4x0 + 4 5.95/1.97 orientation: 5.95/1.97 a(x1) = 4x1 + 4 >= 4x1 = b(b(x1)) 5.95/1.97 5.95/1.97 b(b(a(x1))) = 16x1 + 16 >= 16x1 + 16 = a(c(c(c(b(b(x1)))))) 5.95/1.97 5.95/1.97 b(b(c(c(c(x1))))) = 4x1 + 12 >= 4x1 + 4 = a(x1) 5.95/1.97 problem: 5.95/1.97 b(b(a(x1))) -> a(c(c(c(b(b(x1)))))) 5.95/1.97 KBO Processor: 5.95/1.97 weight function: 5.95/1.97 w0 = 1 5.95/1.97 w(b) = w(a) = 1 5.95/1.97 w(c) = 0 5.95/1.97 precedence: 5.95/1.97 c > b > a 5.95/1.97 problem: 5.95/1.97 5.95/1.97 Qed 5.95/1.97 EOF
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