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SRS_Standard 2019-03-29 03.29 pair #432292254
details
property
value
status
complete
benchmark
z008.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n084.star.cs.uiowa.edu
space
Zantema_04
run statistics
property
value
solver
ttt2-1.19
configuration
ttt2
runtime (wallclock)
1.63282 seconds
cpu usage
5.15388
user time
4.03297
system time
1.12091
max virtual memory
3574952.0
max residence set size
67988.0
stage attributes
key
value
starexec-result
YES
output
3.93/1.62 YES 3.93/1.62 3.93/1.62 Problem: 3.93/1.62 a(s(x1)) -> s(a(x1)) 3.93/1.62 b(a(b(s(x1)))) -> a(b(s(a(x1)))) 3.93/1.62 b(a(b(b(x1)))) -> a(b(a(b(x1)))) 3.93/1.62 a(b(a(a(x1)))) -> b(a(b(a(x1)))) 3.93/1.62 3.93/1.62 Proof: 3.93/1.62 String Reversal Processor: 3.93/1.62 s(a(x1)) -> a(s(x1)) 3.93/1.62 s(b(a(b(x1)))) -> a(s(b(a(x1)))) 3.93/1.62 b(b(a(b(x1)))) -> b(a(b(a(x1)))) 3.93/1.62 a(a(b(a(x1)))) -> a(b(a(b(x1)))) 3.93/1.62 Matrix Interpretation Processor: dim=1 3.93/1.62 3.93/1.62 interpretation: 3.93/1.62 [b](x0) = x0 + 1, 3.93/1.62 3.93/1.62 [a](x0) = x0 + 1, 3.93/1.62 3.93/1.62 [s](x0) = 4x0 + 3 3.93/1.62 orientation: 3.93/1.62 s(a(x1)) = 4x1 + 7 >= 4x1 + 4 = a(s(x1)) 3.93/1.62 3.93/1.62 s(b(a(b(x1)))) = 4x1 + 15 >= 4x1 + 12 = a(s(b(a(x1)))) 3.93/1.62 3.93/1.62 b(b(a(b(x1)))) = x1 + 4 >= x1 + 4 = b(a(b(a(x1)))) 3.93/1.62 3.93/1.62 a(a(b(a(x1)))) = x1 + 4 >= x1 + 4 = a(b(a(b(x1)))) 3.93/1.62 problem: 3.93/1.62 b(b(a(b(x1)))) -> b(a(b(a(x1)))) 3.93/1.62 a(a(b(a(x1)))) -> a(b(a(b(x1)))) 3.93/1.62 Bounds Processor: 3.93/1.62 bound: 0 3.93/1.62 enrichment: match 3.93/1.62 automaton: 3.93/1.62 final states: {6,1} 3.93/1.62 transitions: 3.93/1.62 f30() -> 2* 3.93/1.62 b0(5) -> 1* 3.93/1.62 b0(2) -> 7* 3.93/1.62 b0(8) -> 9* 3.93/1.62 b0(3) -> 4* 3.93/1.62 a0(7) -> 8* 3.93/1.62 a0(2) -> 3* 3.93/1.62 a0(9) -> 6* 3.93/1.62 a0(4) -> 5* 3.93/1.62 1 -> 7* 3.93/1.62 6 -> 3* 3.93/1.62 problem: 3.93/1.62 3.93/1.62 Qed 3.93/1.63 EOF
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