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SRS Standard pair #487091726
details
property
value
status
complete
benchmark
z111.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n145.star.cs.uiowa.edu
space
Zantema_04
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
2.10162 seconds
cpu usage
7.0992
user time
5.46154
system time
1.63766
max virtual memory
5941732.0
max residence set size
66644.0
stage attributes
key
value
starexec-result
YES
output
YES Problem: a(a(x1)) -> b(c(c(c(x1)))) b(c(x1)) -> d(d(d(d(x1)))) a(x1) -> d(c(d(x1))) b(b(x1)) -> c(c(c(x1))) c(c(x1)) -> d(d(d(x1))) c(d(d(x1))) -> a(x1) Proof: Matrix Interpretation Processor: dim=1 interpretation: [c](x0) = x0 + 3, [d](x0) = x0 + 2, [a](x0) = x0 + 7, [b](x0) = x0 + 5 orientation: a(a(x1)) = x1 + 14 >= x1 + 14 = b(c(c(c(x1)))) b(c(x1)) = x1 + 8 >= x1 + 8 = d(d(d(d(x1)))) a(x1) = x1 + 7 >= x1 + 7 = d(c(d(x1))) b(b(x1)) = x1 + 10 >= x1 + 9 = c(c(c(x1))) c(c(x1)) = x1 + 6 >= x1 + 6 = d(d(d(x1))) c(d(d(x1))) = x1 + 7 >= x1 + 7 = a(x1) problem: a(a(x1)) -> b(c(c(c(x1)))) b(c(x1)) -> d(d(d(d(x1)))) a(x1) -> d(c(d(x1))) c(c(x1)) -> d(d(d(x1))) c(d(d(x1))) -> a(x1) String Reversal Processor: a(a(x1)) -> c(c(c(b(x1)))) c(b(x1)) -> d(d(d(d(x1)))) a(x1) -> d(c(d(x1))) c(c(x1)) -> d(d(d(x1))) d(d(c(x1))) -> a(x1) Matrix Interpretation Processor: dim=1 interpretation: [c](x0) = x0 + 2, [d](x0) = x0 + 1, [a](x0) = x0 + 4, [b](x0) = x0 + 2 orientation: a(a(x1)) = x1 + 8 >= x1 + 8 = c(c(c(b(x1)))) c(b(x1)) = x1 + 4 >= x1 + 4 = d(d(d(d(x1)))) a(x1) = x1 + 4 >= x1 + 4 = d(c(d(x1))) c(c(x1)) = x1 + 4 >= x1 + 3 = d(d(d(x1))) d(d(c(x1))) = x1 + 4 >= x1 + 4 = a(x1) problem: a(a(x1)) -> c(c(c(b(x1)))) c(b(x1)) -> d(d(d(d(x1)))) a(x1) -> d(c(d(x1))) d(d(c(x1))) -> a(x1) String Reversal Processor: a(a(x1)) -> b(c(c(c(x1)))) b(c(x1)) -> d(d(d(d(x1)))) a(x1) -> d(c(d(x1))) c(d(d(x1))) -> a(x1) WPO Processor: algebra: Sum weight function: w0 = 0 w(b) = 4 w(a) = 2 w(d) = 1 w(c) = 0 status function: st(d) = st(b) = st(c) = st(a) = [0] precedence: c > a > b > d problem: Qed
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