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TRS Standard pair #487073842
details
property
value
status
complete
benchmark
0.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n186.star.cs.uiowa.edu
space
MNZ_10
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
6.14287 seconds
cpu usage
22.9339
user time
18.8281
system time
4.1058
max virtual memory
6176060.0
max residence set size
171180.0
stage attributes
key
value
starexec-result
YES
output
YES Problem: s(s(0())) -> f(0()) f(0()) -> 0() s(s(s(0()))) -> f(s(0())) f(s(0())) -> s(0()) s(s(s(s(s(s(s(s(0())))))))) -> f(s(s(0()))) f(s(s(0()))) -> s(s(s(s(s(s(0())))))) g(x) -> h(x,x) s(x) -> h(x,0()) s(x) -> h(0(),x) f(g(x)) -> g(g(f(x))) g(s(x)) -> s(s(g(x))) h(f(x),g(x)) -> f(s(x)) Proof: Polynomial Interpretation Processor: dimension: 1 interpretation: [h](x0, x1) = x0 + x1, [s](x0) = x0 + 1, [g](x0) = 4x0 + 6, [0] = 0, [f](x0) = -1x0 + 2x0x0 + 1 orientation: s(s(0())) = 2 >= 1 = f(0()) f(0()) = 1 >= 0 = 0() s(s(s(0()))) = 3 >= 2 = f(s(0())) f(s(0())) = 2 >= 1 = s(0()) s(s(s(s(s(s(s(s(0())))))))) = 8 >= 7 = f(s(s(0()))) f(s(s(0()))) = 7 >= 6 = s(s(s(s(s(s(0())))))) g(x) = 4x + 6 >= 2x = h(x,x) s(x) = x + 1 >= x = h(x,0()) s(x) = x + 1 >= x = h(0(),x) f(g(x)) = 92x + 32x*x + 67 >= -16x + 32x*x + 46 = g(g(f(x))) g(s(x)) = 4x + 10 >= 4x + 8 = s(s(g(x))) h(f(x),g(x)) = 3x + 2x*x + 7 >= 3x + 2x*x + 2 = f(s(x)) problem: Qed
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