details

property	value
status	complete
benchmark	Ex9_Luc06_GM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n189.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	ttt2-1.20
configuration	ttt2
runtime (wallclock)	0.53215 seconds
cpu usage	1.06115
user time	0.75036
system time	0.310787
max virtual memory	96176.0
max residence set size	62604.0

stage attributes

key	value
starexec-result	YES

output

YES Problem: a__f(a(),X,X) -> a__f(X,a__b(),b()) a__b() -> a() mark(f(X1,X2,X3)) -> a__f(X1,mark(X2),X3) mark(b()) -> a__b() mark(a()) -> a() a__f(X1,X2,X3) -> f(X1,X2,X3) a__b() -> b() Proof: Matrix Interpretation Processor: dim=1 interpretation: [f](x0, x1, x2) = 5x0 + x1 + 4x2, [a__f](x0, x1, x2) = 5x0 + x1 + 4x2, [b] = 2, [mark](x0) = x0 + 2, [a] = 2, [a__b] = 2 orientation: a__f(a(),X,X) = 5X + 10 >= 5X + 10 = a__f(X,a__b(),b()) a__b() = 2 >= 2 = a() mark(f(X1,X2,X3)) = 5X1 + X2 + 4X3 + 2 >= 5X1 + X2 + 4X3 + 2 = a__f(X1,mark(X2),X3) mark(b()) = 4 >= 2 = a__b() mark(a()) = 4 >= 2 = a() a__f(X1,X2,X3) = 5X1 + X2 + 4X3 >= 5X1 + X2 + 4X3 = f(X1,X2,X3) a__b() = 2 >= 2 = b() problem: a__f(a(),X,X) -> a__f(X,a__b(),b()) a__b() -> a() mark(f(X1,X2,X3)) -> a__f(X1,mark(X2),X3) a__f(X1,X2,X3) -> f(X1,X2,X3) a__b() -> b() Matrix Interpretation Processor: dim=3 interpretation: [1 1 1] [1 0 0] [1 1 1] [f](x0, x1, x2) = [0 1 0]x0 + [1 0 0]x1 + [0 0 0]x2 [1 0 0] [1 0 0] [0 1 0] , [1 1 1] [1 0 0] [1 1 1] [0] [a__f](x0, x1, x2) = [1 1 0]x0 + [1 0 0]x1 + [0 1 0]x2 + [1] [1 1 0] [1 0 0] [0 1 0] [1], [0] [b] = [0] [1], [1 0 0] [0] [mark](x0) = [1 1 0]x0 + [1] [0 1 1] [1], [0] [a] = [1] [1], [1] [a__b] = [1] [1] orientation: [2 1 1] [2] [1 1 1] [2] a__f(a(),X,X) = [1 1 0]X + [2] >= [1 1 0]X + [2] = a__f(X,a__b(),b()) [1 1 0] [2] [1 1 0] [2] [1] [0] a__b() = [1] >= [1] = a() [1] [1] [1 1 1] [1 0 0] [1 1 1] [0] [1 1 1] [1 0 0] [1 1 1] [0] mark(f(X1,X2,X3)) = [1 2 1]X1 + [2 0 0]X2 + [1 1 1]X3 + [1] >= [1 1 0]X1 + [1 0 0]X2 + [0 1 0]X3 + [1] = a__f(X1,mark(X2),X3) [1 1 0] [2 0 0] [0 1 0] [1] [1 1 0] [1 0 0] [0 1 0] [1] [1 1 1] [1 0 0] [1 1 1] [0] [1 1 1] [1 0 0] [1 1 1] a__f(X1,X2,X3) = [1 1 0]X1 + [1 0 0]X2 + [0 1 0]X3 + [1] >= [0 1 0]X1 + [1 0 0]X2 + [0 0 0]X3 = f(X1,X2,X3) [1 1 0] [1 0 0] [0 1 0] [1] [1 0 0] [1 0 0] [0 1 0] [1] [0] a__b() = [1] >= [0] = b() [1] [1] problem: a__f(a(),X,X) -> a__f(X,a__b(),b()) mark(f(X1,X2,X3)) -> a__f(X1,mark(X2),X3) a__f(X1,X2,X3) -> f(X1,X2,X3) Matrix Interpretation Processor: dim=1 interpretation: [f](x0, x1, x2) = x0 + 4x1 + x2 + 2, popout

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

job log

popout

actions

all output return to TRS Standard