details

property	value
status	complete
benchmark	3.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n072.star.cs.uiowa.edu
space	MNZ_10

run statistics

property	value
solver	ttt2-1.20
configuration	ttt2
runtime (wallclock)	7.87032 seconds
cpu usage	29.6955
user time	24.5359
system time	5.15966
max virtual memory	6195352.0
max residence set size	157396.0

stage attributes

key	value
starexec-result	YES

output

YES Problem: s(s(0())) -> f(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)) s(s(s(0()))) -> k(0()) k(0()) -> s(0()) s(s(s(s(0())))) -> k(s(0())) k(s(0())) -> s(s(0())) s(s(s(s(s(s(s(s(s(0()))))))))) -> k(s(s(0()))) k(s(s(0()))) -> s(s(s(s(s(s(s(0()))))))) h(k(x),g(x)) -> k(s(x)) Proof: Polynomial Interpretation Processor: dimension: 1 interpretation: [h](x0, x1) = x0 + x1, [s](x0) = x0 + 1, [g](x0) = 4x0 + 6, [k](x0) = -1x0 + 2x0x0 + 2, [0] = 0, [f](x0) = x0x0 orientation: s(s(0())) = 2 >= 1 = f(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)) = 48x + 16x*x + 36 >= 16x*x + 30 = g(g(f(x))) g(s(x)) = 4x + 10 >= 4x + 8 = s(s(g(x))) h(f(x),g(x)) = 4x + x*x + 6 >= 2x + x*x + 1 = f(s(x)) s(s(s(0()))) = 3 >= 2 = k(0()) k(0()) = 2 >= 1 = s(0()) s(s(s(s(0())))) = 4 >= 3 = k(s(0())) k(s(0())) = 3 >= 2 = s(s(0())) s(s(s(s(s(s(s(s(s(0()))))))))) = 9 >= 8 = k(s(s(0()))) k(s(s(0()))) = 8 >= 7 = s(s(s(s(s(s(s(0()))))))) h(k(x),g(x)) = 3x + 2x*x + 8 >= 3x + 2x*x + 3 = k(s(x)) problem: Qed popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to TRS Standard