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SRS Standard pair #487091606
details
property
value
status
complete
benchmark
z022.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n137.star.cs.uiowa.edu
space
Zantema_04
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
1.29042 seconds
cpu usage
4.02123
user time
3.15567
system time
0.86556
max virtual memory
5927244.0
max residence set size
68504.0
stage attributes
key
value
starexec-result
YES
output
YES Problem: a(a(b(x1))) -> b(a(b(c(a(x1))))) b(a(x1)) -> a(b(b(x1))) b(c(a(x1))) -> c(a(b(x1))) Proof: String Reversal Processor: b(a(a(x1))) -> a(c(b(a(b(x1))))) a(b(x1)) -> b(b(a(x1))) a(c(b(x1))) -> b(a(c(x1))) Matrix Interpretation Processor: dim=3 interpretation: [1 0 1] [0] [a](x0) = [0 0 0]x0 + [0] [1 1 1] [1], [b](x0) = x0 , [1 0 0] [0] [c](x0) = [0 0 0]x0 + [1] [1 0 0] [0] orientation: [2 1 2] [1] [2 0 2] [0] b(a(a(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = a(c(b(a(b(x1))))) [2 1 2] [2] [2 0 2] [2] [1 0 1] [0] [1 0 1] [0] a(b(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = b(b(a(x1))) [1 1 1] [1] [1 1 1] [1] [2 0 0] [0] [2 0 0] [0] a(c(b(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = b(a(c(x1))) [2 0 0] [2] [2 0 0] [2] problem: a(b(x1)) -> b(b(a(x1))) a(c(b(x1))) -> b(a(c(x1))) String Reversal Processor: b(a(x1)) -> a(b(b(x1))) b(c(a(x1))) -> c(a(b(x1))) KBO Processor: weight function: w0 = 1 w(c) = w(a) = 1 w(b) = 0 precedence: b > a > c problem: Qed
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