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SRS Standard pair #487085588
details
property
value
status
complete
benchmark
size-12-alpha-3-num-92.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n142.star.cs.uiowa.edu
space
Waldmann_07_size12
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
44.6026 seconds
cpu usage
174.366
user time
170.8
system time
3.56692
max virtual memory
1.1833148E7
max residence set size
132844.0
stage attributes
key
value
starexec-result
YES
output
YES Problem: a(x1) -> x1 a(a(x1)) -> b(c(x1)) a(b(b(x1))) -> b(b(a(a(x1)))) Proof: DP Processor: DPs: a#(b(b(x1))) -> a#(x1) a#(b(b(x1))) -> a#(a(x1)) TRS: a(x1) -> x1 a(a(x1)) -> b(c(x1)) a(b(b(x1))) -> b(b(a(a(x1)))) Arctic Interpretation Processor: dimension: 2 usable rules: a(x1) -> x1 a(a(x1)) -> b(c(x1)) a(b(b(x1))) -> b(b(a(a(x1)))) interpretation: [0 -&] [c](x0) = [0 -&]x0, [a#](x0) = [0 0]x0, [0 -&] [-&] [a](x0) = [2 0 ]x0 + [0 ], [0 0] [-&] [b](x0) = [2 0]x0 + [0 ] orientation: a#(b(b(x1))) = [2 2]x1 + [0] >= [0 0]x1 = a#(x1) a#(b(b(x1))) = [2 2]x1 + [0] >= [2 0]x1 + [0] = a#(a(x1)) [0 -&] [-&] a(x1) = [2 0 ]x1 + [0 ] >= x1 = x1 [0 -&] [-&] [0 -&] [-&] a(a(x1)) = [2 0 ]x1 + [0 ] >= [2 -&]x1 + [0 ] = b(c(x1)) [2 0] [0] [2 0] [0] a(b(b(x1))) = [4 2]x1 + [2] >= [4 2]x1 + [2] = b(b(a(a(x1)))) problem: DPs: a#(b(b(x1))) -> a#(a(x1)) TRS: a(x1) -> x1 a(a(x1)) -> b(c(x1)) a(b(b(x1))) -> b(b(a(a(x1)))) Restore Modifier: DPs: a#(b(b(x1))) -> a#(a(x1)) TRS: a(x1) -> x1 a(a(x1)) -> b(c(x1)) a(b(b(x1))) -> b(b(a(a(x1)))) Arctic Interpretation Processor: dimension: 2 usable rules: a(x1) -> x1 a(a(x1)) -> b(c(x1)) a(b(b(x1))) -> b(b(a(a(x1)))) interpretation: [-& -&] [0] [c](x0) = [0 0 ]x0 + [1], [a#](x0) = [-& 0 ]x0 + [0], [0 0 ] [0] [a](x0) = [-& 0 ]x0 + [3], [0 0 ] [3] [b](x0) = [2 -&]x0 + [0] orientation: a#(b(b(x1))) = [2 2]x1 + [5] >= [-& 0 ]x1 + [3] = a#(a(x1)) [0 0 ] [0] a(x1) = [-& 0 ]x1 + [3] >= x1 = x1 [0 0 ] [3] [0 0 ] [3] a(a(x1)) = [-& 0 ]x1 + [3] >= [-& -&]x1 + [2] = b(c(x1)) [2 2] [5] [2 2] [5] a(b(b(x1))) = [2 2]x1 + [5] >= [2 2]x1 + [5] = b(b(a(a(x1)))) problem: DPs: TRS: a(x1) -> x1 a(a(x1)) -> b(c(x1)) a(b(b(x1))) -> b(b(a(a(x1)))) Qed
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