details

property	value
status	complete
benchmark	Ex24_Luc06_C.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)	1.37831 seconds
cpu usage	4.29275
user time	3.46588
system time	0.826862
max virtual memory	1963272.0
max residence set size	83492.0

stage attributes

key	value
starexec-result	YES

output

YES Problem: active(f(b(),X,c())) -> mark(f(X,c(),X)) active(c()) -> mark(b()) active(f(X1,X2,X3)) -> f(X1,active(X2),X3) f(X1,mark(X2),X3) -> mark(f(X1,X2,X3)) proper(f(X1,X2,X3)) -> f(proper(X1),proper(X2),proper(X3)) proper(b()) -> ok(b()) proper(c()) -> ok(c()) f(ok(X1),ok(X2),ok(X3)) -> ok(f(X1,X2,X3)) top(mark(X)) -> top(proper(X)) top(ok(X)) -> top(active(X)) Proof: Matrix Interpretation Processor: dim=2 interpretation: [1 1] [mark](x0) = [0 0]x0, [2] [c] = [0], [1 0] [active](x0) = [0 0]x0, [proper](x0) = x0, [1] [b] = [1], [2 0] [1] [top](x0) = [3 0]x0 + [0], [ok](x0) = x0, [2 3] [3 3] [1 0] [f](x0, x1, x2) = [0 0]x0 + [0 0]x1 + [0 0]x2 orientation: [3 3] [7] [3 3] [6] active(f(b(),X,c())) = [0 0]X + [0] >= [0 0]X + [0] = mark(f(X,c(),X)) [2] [2] active(c()) = [0] >= [0] = mark(b()) [2 3] [3 3] [1 0] [2 3] [3 0] [1 0] active(f(X1,X2,X3)) = [0 0]X1 + [0 0]X2 + [0 0]X3 >= [0 0]X1 + [0 0]X2 + [0 0]X3 = f(X1,active(X2),X3) [2 3] [3 3] [1 0] [2 3] [3 3] [1 0] f(X1,mark(X2),X3) = [0 0]X1 + [0 0]X2 + [0 0]X3 >= [0 0]X1 + [0 0]X2 + [0 0]X3 = mark(f(X1,X2,X3)) [2 3] [3 3] [1 0] [2 3] [3 3] [1 0] proper(f(X1,X2,X3)) = [0 0]X1 + [0 0]X2 + [0 0]X3 >= [0 0]X1 + [0 0]X2 + [0 0]X3 = f(proper(X1),proper(X2),proper(X3)) [1] [1] proper(b()) = [1] >= [1] = ok(b()) [2] [2] proper(c()) = [0] >= [0] = ok(c()) [2 3] [3 3] [1 0] [2 3] [3 3] [1 0] f(ok(X1),ok(X2),ok(X3)) = [0 0]X1 + [0 0]X2 + [0 0]X3 >= [0 0]X1 + [0 0]X2 + [0 0]X3 = ok(f(X1,X2,X3)) [2 2] [1] [2 0] [1] top(mark(X)) = [3 3]X + [0] >= [3 0]X + [0] = top(proper(X)) [2 0] [1] [2 0] [1] top(ok(X)) = [3 0]X + [0] >= [3 0]X + [0] = top(active(X)) problem: active(c()) -> mark(b()) active(f(X1,X2,X3)) -> f(X1,active(X2),X3) f(X1,mark(X2),X3) -> mark(f(X1,X2,X3)) proper(f(X1,X2,X3)) -> f(proper(X1),proper(X2),proper(X3)) proper(b()) -> ok(b()) proper(c()) -> ok(c()) f(ok(X1),ok(X2),ok(X3)) -> ok(f(X1,X2,X3)) top(mark(X)) -> top(proper(X)) top(ok(X)) -> top(active(X)) Matrix Interpretation Processor: dim=3 interpretation: [1 1 0] [mark](x0) = [0 1 0]x0 [0 0 1] , [1] [c] = [0] [0], [1 0 0] [active](x0) = [0 0 0]x0 [0 0 1] , [1 0 1] [proper](x0) = [0 0 0]x0 [0 0 0] , popout

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

job log

popout

actions

all output return to TRS Standard