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TRS Standard pair #516965464
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
n137.star.cs.uiowa.edu
space
Zantema_05
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
0.868933916092 seconds
cpu usage
2.086727351
max memory
3.80608512E8
stage attributes
key
value
output-size
5745
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: 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=1 interpretation: [a0] = 0, [b0] = 1, [f](x0, x1) = 5x0 + 3x1, [a1](x0) = 3x0, [c1](x0) = 3x0, [b1](x0) = 3x0, [c0] = 0 orientation: a1(a1(x)) = 9x >= 9x = c1(b1(x)) b1(b1(x)) = 9x >= 9x = a1(c1(x)) c1(c1(x)) = 9x >= 9x = b1(a1(x)) f(a0(),x1) = 3x1 >= 3x1 = a1(x1) f(c0(),x3) = 3x3 >= 3x3 = c1(x3) f(b0(),x5) = 3x5 + 5 >= 3x5 = b1(x5) 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) Matrix Interpretation Processor: dim=1 interpretation: [a0] = 0, [f](x0, x1) = 3x0 + 2x1 + 4, [a1](x0) = 2x0 + 4, [c1](x0) = 2x0 + 4, [b1](x0) = 2x0 + 4, [c0] = 3 orientation: a1(a1(x)) = 4x + 12 >= 4x + 12 = c1(b1(x)) b1(b1(x)) = 4x + 12 >= 4x + 12 = a1(c1(x)) c1(c1(x)) = 4x + 12 >= 4x + 12 = b1(a1(x)) f(a0(),x1) = 2x1 + 4 >= 2x1 + 4 = a1(x1) f(c0(),x3) = 2x3 + 13 >= 2x3 + 4 = c1(x3) problem: a1(a1(x)) -> c1(b1(x)) b1(b1(x)) -> a1(c1(x)) c1(c1(x)) -> b1(a1(x)) f(a0(),x1) -> a1(x1) Matrix Interpretation Processor: dim=3 interpretation:
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