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SRS Standard pair #487091096
details
property
value
status
complete
benchmark
3-matchbox.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n141.star.cs.uiowa.edu
space
Secret_06_SRS
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
3.23962 seconds
cpu usage
22.8892
user time
19.8083
system time
3.0809
max virtual memory
6062808.0
max residence set size
74240.0
stage attributes
key
value
starexec-result
YES
output
YES Problem: c(b(a(a(x1)))) -> a(a(b(c(x1)))) b(a(a(a(x1)))) -> a(a(a(b(x1)))) a(b(c(x1))) -> c(b(a(x1))) c(c(b(b(x1)))) -> b(b(c(c(x1)))) Proof: Matrix Interpretation Processor: dim=2 interpretation: [2 0] [1] [b](x0) = [0 1]x0 + [0], [1 1] [1] [a](x0) = [0 1]x0 + [0], [1 0] [0] [c](x0) = [0 2]x0 + [1] orientation: [2 4] [5] [2 4] [5] c(b(a(a(x1)))) = [0 2]x1 + [1] >= [0 2]x1 + [1] = a(a(b(c(x1)))) [2 6] [7] [2 3] [4] b(a(a(a(x1)))) = [0 1]x1 + [0] >= [0 1]x1 + [0] = a(a(a(b(x1)))) [2 2] [3] [2 2] [3] a(b(c(x1))) = [0 2]x1 + [1] >= [0 2]x1 + [1] = c(b(a(x1))) [4 0] [3] [4 0] [3] c(c(b(b(x1)))) = [0 4]x1 + [3] >= [0 4]x1 + [3] = b(b(c(c(x1)))) problem: c(b(a(a(x1)))) -> a(a(b(c(x1)))) a(b(c(x1))) -> c(b(a(x1))) c(c(b(b(x1)))) -> b(b(c(c(x1)))) String Reversal Processor: a(a(b(c(x1)))) -> c(b(a(a(x1)))) c(b(a(x1))) -> a(b(c(x1))) b(b(c(c(x1)))) -> c(c(b(b(x1)))) Matrix Interpretation Processor: dim=3 interpretation: [1 0 0] [0] [b](x0) = [0 0 1]x0 + [0] [0 1 0] [1], [1 1 0] [a](x0) = [0 1 1]x0 [0 0 0] , [1 0 0] [c](x0) = [0 0 1]x0 [0 1 0] orientation: [1 2 1] [1] [1 2 1] [0] a(a(b(c(x1)))) = [0 1 1]x1 + [1] >= [0 1 1]x1 + [1] = c(b(a(a(x1)))) [0 0 0] [0] [0 0 0] [0] [1 1 0] [0] [1 1 0] [0] c(b(a(x1))) = [0 1 1]x1 + [1] >= [0 1 1]x1 + [1] = a(b(c(x1))) [0 0 0] [0] [0 0 0] [0] [0] [0] b(b(c(c(x1)))) = x1 + [1] >= x1 + [1] = c(c(b(b(x1)))) [1] [1] problem: c(b(a(x1))) -> a(b(c(x1))) b(b(c(c(x1)))) -> c(c(b(b(x1)))) String Reversal Processor: a(b(c(x1))) -> c(b(a(x1))) c(c(b(b(x1)))) -> b(b(c(c(x1)))) KBO Processor: weight function: w0 = 1 w(c) = w(b) = 1 w(a) = 0 precedence: a > c > b problem: Qed
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