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TRS Stand 20472 pair #381712725
details
property
value
status
complete
benchmark
z05.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n191.star.cs.uiowa.edu
space
Zantema_05
run statistics
property
value
solver
ttt2-1.17+nonreach
configuration
ttt2-1.17+nonreach
runtime (wallclock)
0.915951013565 seconds
cpu usage
2.130988137
max memory
2.50658816E8
stage attributes
key
value
output-size
4929
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES Problem: f(a(),f(a(),x)) -> f(c(),f(b(),x)) f(b(),f(b(),x)) -> f(a(),f(c(),x)) f(c(),f(c(),x)) -> f(b(),f(a(),x)) Proof: Extended Uncurrying Processor: application symbol: f symbol table: b ==> b0/0 b1/1 c ==> c0/0 c1/1 a ==> a0/0 a1/1 uncurry-rules: f(a0(),x1) -> a1(x1) f(c0(),x3) -> c1(x3) f(b0(),x5) -> b1(x5) eta-rules: problem: a1(a1(x)) -> c1(b1(x)) b1(b1(x)) -> a1(c1(x)) c1(c1(x)) -> b1(a1(x)) f(a0(),x1) -> a1(x1) f(c0(),x3) -> c1(x3) f(b0(),x5) -> b1(x5) Matrix Interpretation Processor: dim=3 interpretation: [1 1 1] [0] [b1](x0) = [0 0 0]x0 + [0] [0 0 0] [1], [0] [b0] = [1] [0], [1 0 1] [0] [c1](x0) = [0 0 0]x0 + [0] [0 1 0] [1], [0] [c0] = [0] [0], [1 1 1] [0] [a1](x0) = [0 0 0]x0 + [0] [0 0 0] [1], [0] [a0] = [0] [0], [1 0 0] [1 1 1] [1] [f](x0, x1) = [0 1 0]x0 + [0 1 0]x1 + [0] [0 0 0] [0 1 0] [1] orientation: [1 1 1] [1] [1 1 1] [1] a1(a1(x)) = [0 0 0]x + [0] >= [0 0 0]x + [0] = c1(b1(x)) [0 0 0] [1] [0 0 0] [1] [1 1 1] [1] [1 1 1] [1] b1(b1(x)) = [0 0 0]x + [0] >= [0 0 0]x + [0] = a1(c1(x)) [0 0 0] [1] [0 0 0] [1] [1 1 1] [1] [1 1 1] [1] c1(c1(x)) = [0 0 0]x + [0] >= [0 0 0]x + [0] = b1(a1(x)) [0 0 0] [1] [0 0 0] [1] [1 1 1] [1] [1 1 1] [0] f(a0(),x1) = [0 1 0]x1 + [0] >= [0 0 0]x1 + [0] = a1(x1) [0 1 0] [1] [0 0 0] [1] [1 1 1] [1] [1 0 1] [0] f(c0(),x3) = [0 1 0]x3 + [0] >= [0 0 0]x3 + [0] = c1(x3) [0 1 0] [1] [0 1 0] [1] [1 1 1] [1] [1 1 1] [0] f(b0(),x5) = [0 1 0]x5 + [1] >= [0 0 0]x5 + [0] = b1(x5) [0 1 0] [1] [0 0 0] [1] problem: a1(a1(x)) -> c1(b1(x)) b1(b1(x)) -> a1(c1(x)) c1(c1(x)) -> b1(a1(x)) DP Processor: DPs: a{1,#}(a1(x)) -> b{1,#}(x) a{1,#}(a1(x)) -> c{1,#}(b1(x)) b{1,#}(b1(x)) -> c{1,#}(x) b{1,#}(b1(x)) -> a{1,#}(c1(x)) c{1,#}(c1(x)) -> a{1,#}(x) c{1,#}(c1(x)) -> b{1,#}(a1(x))
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