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TRS Standard pair #516965714
details
property
value
status
complete
benchmark
z07.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n028.star.cs.uiowa.edu
space
Zantema_05
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
0.691381931305 seconds
cpu usage
0.858090709
max memory
2.08744448E8
stage attributes
key
value
output-size
5117
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_ttt2 /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES Problem: f(a(),x) -> f(b(),f(c(),x)) f(a(),f(b(),x)) -> f(b(),f(a(),x)) f(d(),f(c(),x)) -> f(d(),f(a(),x)) f(a(),f(c(),x)) -> f(c(),f(a(),x)) Proof: Extended Uncurrying Processor: application symbol: f symbol table: d ==> d0/0 d1/1 c ==> c0/0 c1/1 b ==> b0/0 b1/1 a ==> a0/0 a1/1 uncurry-rules: f(a0(),x1) -> a1(x1) f(b0(),x3) -> b1(x3) f(c0(),x5) -> c1(x5) f(d0(),x7) -> d1(x7) eta-rules: problem: a1(x) -> b1(c1(x)) a1(b1(x)) -> b1(a1(x)) d1(c1(x)) -> d1(a1(x)) a1(c1(x)) -> c1(a1(x)) f(a0(),x1) -> a1(x1) f(b0(),x3) -> b1(x3) f(c0(),x5) -> c1(x5) f(d0(),x7) -> d1(x7) Matrix Interpretation Processor: dim=3 interpretation: [1 0 0] [b1](x0) = [0 0 0]x0 [0 0 0] , [1 0 0] [1 1 0] [1] [f](x0, x1) = [0 0 1]x0 + [0 1 0]x1 + [0] [0 0 0] [0 0 0] [0], [0] [a0] = [0] [0], [1 0 0] [0] [c1](x0) = [0 1 0]x0 + [1] [0 0 0] [0], [0] [b0] = [0] [0], [1 1 0] [d1](x0) = [0 1 0]x0 [0 0 0] , [0] [d0] = [0] [0], [0] [c0] = [0] [1], [1 0 0] [1] [a1](x0) = [0 1 0]x0 + [0] [0 0 0] [0] orientation: [1 0 0] [1] [1 0 0] a1(x) = [0 1 0]x + [0] >= [0 0 0]x = b1(c1(x)) [0 0 0] [0] [0 0 0] [1 0 0] [1] [1 0 0] [1] a1(b1(x)) = [0 0 0]x + [0] >= [0 0 0]x + [0] = b1(a1(x)) [0 0 0] [0] [0 0 0] [0] [1 1 0] [1] [1 1 0] [1] d1(c1(x)) = [0 1 0]x + [1] >= [0 1 0]x + [0] = d1(a1(x)) [0 0 0] [0] [0 0 0] [0] [1 0 0] [1] [1 0 0] [1] a1(c1(x)) = [0 1 0]x + [1] >= [0 1 0]x + [1] = c1(a1(x)) [0 0 0] [0] [0 0 0] [0] [1 1 0] [1] [1 0 0] [1] f(a0(),x1) = [0 1 0]x1 + [0] >= [0 1 0]x1 + [0] = a1(x1) [0 0 0] [0] [0 0 0] [0] [1 1 0] [1] [1 0 0] f(b0(),x3) = [0 1 0]x3 + [0] >= [0 0 0]x3 = b1(x3)
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