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SRS Standard pair #516976316
details
property
value
status
complete
benchmark
random-470.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n010.star.cs.uiowa.edu
space
Waldmann_19
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
5.02372598648 seconds
cpu usage
18.732738113
max memory
9.51721984E8
stage attributes
key
value
output-size
3124
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_ttt2 /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES Problem: a(b(a(a(x1)))) -> a(b(b(a(x1)))) a(b(a(b(x1)))) -> a(b(b(a(x1)))) b(b(a(a(x1)))) -> a(b(a(a(x1)))) b(b(b(b(x1)))) -> a(a(a(a(x1)))) Proof: String Reversal Processor: a(a(b(a(x1)))) -> a(b(b(a(x1)))) b(a(b(a(x1)))) -> a(b(b(a(x1)))) a(a(b(b(x1)))) -> a(a(b(a(x1)))) b(b(b(b(x1)))) -> a(a(a(a(x1)))) Matrix Interpretation Processor: dim=3 interpretation: [1 0 1] [0] [b](x0) = [1 0 1]x0 + [1] [0 1 0] [0], [1 1 0] [a](x0) = [0 0 0]x0 [0 0 0] orientation: [2 2 0] [1] [2 2 0] [1] a(a(b(a(x1)))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = a(b(b(a(x1)))) [0 0 0] [0] [0 0 0] [0] [2 2 0] [1] [2 2 0] [1] b(a(b(a(x1)))) = [2 2 0]x1 + [2] >= [0 0 0]x1 + [0] = a(b(b(a(x1)))) [0 0 0] [0] [0 0 0] [0] [2 2 2] [1] [2 2 0] [1] a(a(b(b(x1)))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = a(a(b(a(x1)))) [0 0 0] [0] [0 0 0] [0] [3 2 3] [2] [1 1 0] b(b(b(b(x1)))) = [3 2 3]x1 + [3] >= [0 0 0]x1 = a(a(a(a(x1)))) [2 1 2] [2] [0 0 0] problem: a(a(b(a(x1)))) -> a(b(b(a(x1)))) b(a(b(a(x1)))) -> a(b(b(a(x1)))) a(a(b(b(x1)))) -> a(a(b(a(x1)))) String Reversal Processor: a(b(a(a(x1)))) -> a(b(b(a(x1)))) a(b(a(b(x1)))) -> a(b(b(a(x1)))) b(b(a(a(x1)))) -> a(b(a(a(x1)))) Matrix Interpretation Processor: dim=3 interpretation: [1 0 1] [1] [b](x0) = [1 0 2]x0 + [0] [0 1 0] [0], [1 0 0] [1] [a](x0) = [0 0 0]x0 + [0] [1 0 0] [0] orientation: [2 0 0] [5] [2 0 0] [4] a(b(a(a(x1)))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = a(b(b(a(x1)))) [2 0 0] [4] [2 0 0] [3] [2 0 2] [5] [2 0 0] [4] a(b(a(b(x1)))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = a(b(b(a(x1)))) [2 0 2] [4] [2 0 0] [3] [2 0 0] [5] [2 0 0] [5] b(b(a(a(x1)))) = [2 0 0]x1 + [4] >= [0 0 0]x1 + [0] = a(b(a(a(x1)))) [3 0 0] [4] [2 0 0] [4] problem: b(b(a(a(x1)))) -> a(b(a(a(x1)))) KBO Processor: weight function: w0 = 1 w(b) = w(a) = 1 precedence: b > a problem: Qed
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