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SRS Standard pair #487092032
details
property
value
status
complete
benchmark
un06.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n143.star.cs.uiowa.edu
space
Trafo_06
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
5.517 seconds
cpu usage
20.5721
user time
18.8421
system time
1.73001
max virtual memory
6021292.0
max residence set size
98540.0
stage attributes
key
value
starexec-result
YES
output
YES Problem: b(b(x1)) -> b(a(b(a(a(b(x1)))))) b(b(a(b(x1)))) -> b(a(b(b(x1)))) b(a(a(a(b(a(a(b(x1)))))))) -> b(a(a(a(b(b(x1)))))) b(b(a(a(b(x1))))) -> b(a(a(b(b(x1))))) Proof: Matrix Interpretation Processor: dim=4 interpretation: [1 0 0 1] [0] [0 0 1 0] [1] [a](x0) = [0 0 0 0]x0 + [0] [0 1 0 0] [0], [1 0 0 0] [0 0 0 0] [b](x0) = [0 0 1 1]x0 [0 0 0 0] orientation: [1 0 0 0] [1 0 0 0] [0 0 0 0] [0 0 0 0] b(b(x1)) = [0 0 1 1]x1 >= [0 0 0 0]x1 = b(a(b(a(a(b(x1)))))) [0 0 0 0] [0 0 0 0] [1 0 0 0] [1 0 0 0] [0 0 0 0] [0 0 0 0] b(b(a(b(x1)))) = [0 0 0 0]x1 >= [0 0 0 0]x1 = b(a(b(b(x1)))) [0 0 0 0] [0 0 0 0] [1 0 1 1] [2] [1 0 1 1] [1] [0 0 0 0] [0] [0 0 0 0] [0] b(a(a(a(b(a(a(b(x1)))))))) = [0 0 0 0]x1 + [1] >= [0 0 0 0]x1 + [1] = b(a(a(a(b(b(x1)))))) [0 0 0 0] [0] [0 0 0 0] [0] [1 0 0 0] [0] [1 0 0 0] [0] [0 0 0 0] [0] [0 0 0 0] [0] b(b(a(a(b(x1))))) = [0 0 1 1]x1 + [1] >= [0 0 1 1]x1 + [1] = b(a(a(b(b(x1))))) [0 0 0 0] [0] [0 0 0 0] [0] problem: b(b(x1)) -> b(a(b(a(a(b(x1)))))) b(b(a(b(x1)))) -> b(a(b(b(x1)))) b(b(a(a(b(x1))))) -> b(a(a(b(b(x1))))) Matrix Interpretation Processor: dim=3 interpretation: [1 0 0] [a](x0) = [0 0 0]x0 [1 0 1] , [1 1 0] [0] [b](x0) = [0 0 0]x0 + [1] [0 1 0] [0] orientation: [1 1 0] [1] [1 1 0] [0] b(b(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = b(a(b(a(a(b(x1)))))) [0 0 0] [1] [0 0 0] [0] [1 1 0] [1] [1 1 0] [1] b(b(a(b(x1)))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = b(a(b(b(x1)))) [0 0 0] [1] [0 0 0] [0] [1 1 0] [1] [1 1 0] [1] b(b(a(a(b(x1))))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = b(a(a(b(b(x1))))) [0 0 0] [1] [0 0 0] [0] problem: b(b(a(b(x1)))) -> b(a(b(b(x1)))) b(b(a(a(b(x1))))) -> b(a(a(b(b(x1))))) String Reversal Processor: b(a(b(b(x1)))) -> b(b(a(b(x1)))) b(a(a(b(b(x1))))) -> b(b(a(a(b(x1))))) KBO Processor: weight function: w0 = 1 w(b) = 1 w(a) = 0 precedence: a > b problem: Qed
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