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SRS Stand 10685 pair #381718315
details
property
value
status
complete
benchmark
multum3.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n083.star.cs.uiowa.edu
space
Secret_06_SRS
run statistics
property
value
solver
ttt2-1.17+nonreach
configuration
ttt2-1.17+nonreach
runtime (wallclock)
27.8318948746 seconds
cpu usage
109.875327372
max memory
1.667719168E9
stage attributes
key
value
output-size
2510
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES Problem: a(a(b(x1))) -> b(b(a(a(x1)))) b(a(b(x1))) -> a(a(a(a(x1)))) Proof: String Reversal Processor: b(a(a(x1))) -> a(a(b(b(x1)))) b(a(b(x1))) -> a(a(a(a(x1)))) DP Processor: DPs: b#(a(a(x1))) -> b#(x1) b#(a(a(x1))) -> b#(b(x1)) TRS: b(a(a(x1))) -> a(a(b(b(x1)))) b(a(b(x1))) -> a(a(a(a(x1)))) Arctic Interpretation Processor: dimension: 2 usable rules: b(a(a(x1))) -> a(a(b(b(x1)))) b(a(b(x1))) -> a(a(a(a(x1)))) interpretation: [b#](x0) = [1 0]x0, [-& 1 ] [0 ] [a](x0) = [0 -&]x0 + [-&], [0 -&] [-&] [b](x0) = [2 0 ]x0 + [0 ] orientation: b#(a(a(x1))) = [2 1]x1 + [1] >= [1 0]x1 = b#(x1) b#(a(a(x1))) = [2 1]x1 + [1] >= [2 0]x1 + [0] = b#(b(x1)) [1 -&] [0] [1 -&] [0] b(a(a(x1))) = [3 1 ]x1 + [2] >= [3 1 ]x1 + [1] = a(a(b(b(x1)))) [3 1] [1] [2 -&] [1] b(a(b(x1))) = [5 3]x1 + [3] >= [-& 2 ]x1 + [1] = a(a(a(a(x1)))) problem: DPs: b#(a(a(x1))) -> b#(b(x1)) TRS: b(a(a(x1))) -> a(a(b(b(x1)))) b(a(b(x1))) -> a(a(a(a(x1)))) Restore Modifier: DPs: b#(a(a(x1))) -> b#(b(x1)) TRS: b(a(a(x1))) -> a(a(b(b(x1)))) b(a(b(x1))) -> a(a(a(a(x1)))) Arctic Interpretation Processor: dimension: 2 usable rules: b(a(a(x1))) -> a(a(b(b(x1)))) b(a(b(x1))) -> a(a(a(a(x1)))) interpretation: [b#](x0) = [-& 0 ]x0 + [0], [-& 0 ] [0] [a](x0) = [1 -&]x0 + [1], [0 3 ] [1 ] [b](x0) = [-& 0 ]x0 + [-&] orientation: b#(a(a(x1))) = [-& 1 ]x1 + [1] >= [-& 0 ]x1 + [0] = b#(b(x1)) [1 4 ] [4] [1 4 ] [2] b(a(a(x1))) = [-& 1 ]x1 + [1] >= [-& 1 ]x1 + [1] = a(a(b(b(x1)))) [4 7] [5] [2 -&] [2] b(a(b(x1))) = [1 4]x1 + [2] >= [-& 2 ]x1 + [2] = a(a(a(a(x1)))) problem: DPs: TRS: b(a(a(x1))) -> a(a(b(b(x1)))) b(a(b(x1))) -> a(a(a(a(x1)))) Qed
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