details

property	value
status	complete
benchmark	2.44.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n030.star.cs.uiowa.edu
space	SK90

run statistics

property	value
solver	ttt2-1.20
configuration	ttt2
runtime (wallclock)	0.735842943192 seconds
cpu usage	1.314313806
max memory	2.1829632E8

stage attributes

key	value
output-size	3006
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: del(.(x,.(y,z))) -> f(=(x,y),x,y,z) f(true(),x,y,z) -> del(.(y,z)) f(false(),x,y,z) -> .(x,del(.(y,z))) =(nil(),nil()) -> true() =(.(x,y),nil()) -> false() =(nil(),.(y,z)) -> false() =(.(x,y),.(u(),v())) -> and(=(x,u()),=(y,v())) Proof: Matrix Interpretation Processor: dim=2 interpretation: [0] [true] = [0], [0] [v] = [2], [1 1] [del](x0) = [0 2]x0, [1 0] [2 0] [3 0] [2 0] [3] [f](x0, x1, x2, x3) = [1 0]x0 + [1 0]x1 + [3 0]x2 + [2 0]x3 + [2], [0] [false] = [0], [2 0] [1 0] [1] [.](x0, x1) = [1 0]x0 + [1 0]x1 + [0], [0] [u] = [0], [2 0] [1 1] [and](x0, x1) = [0 0]x0 + [0 0]x1, [0] [nil] = [3], [1 0] [=](x0, x1) = [0 0]x0 + x1 orientation: [3 0] [4 0] [2 0] [3] [3 0] [4 0] [2 0] [3] del(.(x,.(y,z))) = [2 0]x + [4 0]y + [2 0]z + [2] >= [2 0]x + [4 0]y + [2 0]z + [2] = f(=(x,y),x,y,z) [2 0] [3 0] [2 0] [3] [3 0] [2 0] [1] f(true(),x,y,z) = [1 0]x + [3 0]y + [2 0]z + [2] >= [2 0]y + [2 0]z + [0] = del(.(y,z)) [2 0] [3 0] [2 0] [3] [2 0] [3 0] [2 0] [2] f(false(),x,y,z) = [1 0]x + [3 0]y + [2 0]z + [2] >= [1 0]x + [3 0]y + [2 0]z + [1] = .(x,del(.(y,z))) [0] [0] =(nil(),nil()) = [3] >= [0] = true() [2 0] [1 0] [1] [0] =(.(x,y),nil()) = [0 0]x + [0 0]y + [3] >= [0] = false() [2 0] [1 0] [1] [0] =(nil(),.(y,z)) = [1 0]y + [1 0]z + [0] >= [0] = false() [2 0] [1 0] [2] [2 0] [1 0] [2] =(.(x,y),.(u(),v())) = [0 0]x + [0 0]y + [0] >= [0 0]x + [0 0]y + [0] = and(=(x,u()),=(y,v())) problem: del(.(x,.(y,z))) -> f(=(x,y),x,y,z) =(nil(),nil()) -> true() =(.(x,y),.(u(),v())) -> and(=(x,u()),=(y,v())) Matrix Interpretation Processor: dim=1 interpretation: [true] = 0, [v] = 2, [del](x0) = 7x0 + 4, [f](x0, x1, x2, x3) = 4x0 + 4x1 + 4x2 + 4x3, [.](x0, x1) = 3x0 + x1, [u] = 1, [and](x0, x1) = x0 + x1, [nil] = 4, [=](x0, x1) = 4x0 + 2x1 orientation: del(.(x,.(y,z))) = 21x + 21y + 7z + 4 >= 20x + 12y + 4z = f(=(x,y),x,y,z) =(nil(),nil()) = 24 >= 0 = true() popout

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