details

property	value
status	complete
benchmark	24.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n180.star.cs.uiowa.edu
space	Various_04

run statistics

property	value
solver	ttt2-1.20
configuration	ttt2
runtime (wallclock)	0.690244 seconds
cpu usage	1.64308
user time	1.26254
system time	0.380547
max virtual memory	1552064.0
max residence set size	68812.0

stage attributes

key	value
starexec-result	YES

output

YES Problem: max(L(x)) -> x max(N(L(0()),L(y))) -> y max(N(L(s(x)),L(s(y)))) -> s(max(N(L(x),L(y)))) max(N(L(x),N(y,z))) -> max(N(L(x),L(max(N(y,z))))) Proof: Matrix Interpretation Processor: dim=3 interpretation: [1 0 0] [s](x0) = [0 0 0]x0 [0 0 0] , [max](x0) = x0 , [1 0 0] [N](x0, x1) = [0 0 0]x0 + x1 [0 1 0] , [L](x0) = x0 , [1] [0] = [1] [0] orientation: max(L(x)) = x >= x = x [1] max(N(L(0()),L(y))) = y + [0] >= y = y [1] [1 0 0] [1 0 0] [1 0 0] [1 0 0] max(N(L(s(x)),L(s(y)))) = [0 0 0]x + [0 0 0]y >= [0 0 0]x + [0 0 0]y = s(max(N(L(x),L(y)))) [0 0 0] [0 0 0] [0 0 0] [0 0 0] [1 0 0] [1 0 0] [1 0 0] [1 0 0] max(N(L(x),N(y,z))) = [0 0 0]x + [0 0 0]y + z >= [0 0 0]x + [0 0 0]y + z = max(N(L(x),L(max(N(y,z))))) [0 1 0] [0 1 0] [0 1 0] [0 1 0] problem: max(L(x)) -> x max(N(L(s(x)),L(s(y)))) -> s(max(N(L(x),L(y)))) max(N(L(x),N(y,z))) -> max(N(L(x),L(max(N(y,z))))) Matrix Interpretation Processor: dim=3 interpretation: [0] [s](x0) = x0 + [1] [0], [1 0 0] [0] [max](x0) = [1 0 1]x0 + [0] [0 0 1] [1], [1 0 1] [1 1 0] [0] [N](x0, x1) = [0 0 0]x0 + [0 0 0]x1 + [1] [0 0 0] [0 0 0] [0], [1 0 0] [L](x0) = [0 0 0]x0 [0 1 1] orientation: [1 0 0] [0] max(L(x)) = [1 1 1]x + [0] >= x = x [0 1 1] [1] [1 1 1] [1 0 0] [1] [1 1 1] [1 0 0] [0] max(N(L(s(x)),L(s(y)))) = [1 1 1]x + [1 0 0]y + [1] >= [1 1 1]x + [1 0 0]y + [1] = s(max(N(L(x),L(y)))) [0 0 0] [0 0 0] [1] [0 0 0] [0 0 0] [1] [1 1 1] [1 0 1] [1 1 0] [1] [1 1 1] [1 0 1] [1 1 0] [0] max(N(L(x),N(y,z))) = [1 1 1]x + [1 0 1]y + [1 1 0]z + [1] >= [1 1 1]x + [1 0 1]y + [1 1 0]z + [0] = max(N(L(x),L(max(N(y,z))))) [0 0 0] [0 0 0] [0 0 0] [1] [0 0 0] [0 0 0] [0 0 0] [1] problem: max(L(x)) -> x Matrix Interpretation Processor: dim=3 interpretation: [1] [max](x0) = x0 + [0] [0], [0] [L](x0) = x0 + [1] [0] orientation: [1] max(L(x)) = x + [1] >= x = x [0] problem: Qed popout

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