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TRS Standard pair #487068462
details
property
value
status
complete
benchmark
z25.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n070.star.cs.uiowa.edu
space
Zantema_05
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
0.563885 seconds
cpu usage
1.17297
user time
0.838973
system time
0.333994
max virtual memory
96176.0
max residence set size
67604.0
stage attributes
key
value
starexec-result
YES
output
YES Problem: f(a(),f(b(),x)) -> f(b(),f(a(),x)) f(b(),f(c(),x)) -> f(c(),f(b(),x)) f(c(),f(a(),x)) -> f(a(),f(c(),x)) Proof: Extended Uncurrying Processor: application symbol: f symbol table: 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) eta-rules: problem: a1(b1(x)) -> b1(a1(x)) b1(c1(x)) -> c1(b1(x)) c1(a1(x)) -> a1(c1(x)) f(a0(),x1) -> a1(x1) f(b0(),x3) -> b1(x3) f(c0(),x5) -> c1(x5) Matrix Interpretation Processor: dim=1 interpretation: [a0] = 1, [c0] = 2, [a1](x0) = 2x0 + 2, [b1](x0) = 2x0 + 2, [c1](x0) = 2x0 + 2, [b0] = 1, [f](x0, x1) = x0 + 2x1 + 1 orientation: a1(b1(x)) = 4x + 6 >= 4x + 6 = b1(a1(x)) b1(c1(x)) = 4x + 6 >= 4x + 6 = c1(b1(x)) c1(a1(x)) = 4x + 6 >= 4x + 6 = a1(c1(x)) f(a0(),x1) = 2x1 + 2 >= 2x1 + 2 = a1(x1) f(b0(),x3) = 2x3 + 2 >= 2x3 + 2 = b1(x3) f(c0(),x5) = 2x5 + 3 >= 2x5 + 2 = c1(x5) problem: a1(b1(x)) -> b1(a1(x)) b1(c1(x)) -> c1(b1(x)) c1(a1(x)) -> a1(c1(x)) f(a0(),x1) -> a1(x1) f(b0(),x3) -> b1(x3) Matrix Interpretation Processor: dim=3 interpretation: [0] [a0] = [0] [1], [1 0 1] [0] [a1](x0) = [0 0 0]x0 + [1] [0 1 0] [0], [1 0 1] [0] [b1](x0) = [0 0 0]x0 + [1] [0 1 0] [0], [1 1 1] [0] [c1](x0) = [0 0 0]x0 + [1] [0 0 0] [1], [0] [b0] = [0] [0], [1 0 1] [1 0 1] [0] [f](x0, x1) = [0 0 0]x0 + [1 0 0]x1 + [1] [0 0 0] [1 1 0] [0] orientation: [1 1 1] [0] [1 1 1] [0] a1(b1(x)) = [0 0 0]x + [1] >= [0 0 0]x + [1] = b1(a1(x)) [0 0 0] [1] [0 0 0] [1] [1 1 1] [1] [1 1 1] [1] b1(c1(x)) = [0 0 0]x + [1] >= [0 0 0]x + [1] = c1(b1(x)) [0 0 0] [1] [0 0 0] [1] [1 1 1] [1] [1 1 1] [1] c1(a1(x)) = [0 0 0]x + [1] >= [0 0 0]x + [1] = a1(c1(x)) [0 0 0] [1] [0 0 0] [1]
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