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TRS Stand 20472 pair #381712127
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
n192.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
ttt2-1.17+nonreach
configuration
ttt2-1.17+nonreach
runtime (wallclock)
1.86510109901 seconds
cpu usage
4.963692487
max memory
2.96292352E8
stage attributes
key
value
output-size
7594
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- 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: [2 0] [2] [top](x0) = [1 1]x0 + [0], [ok](x0) = x0, [proper](x0) = x0, [1 1] [mark](x0) = [0 0]x0, [1 0] [active](x0) = [0 0]x0, [1 2] [3 2] [2 0] [f](x0, x1, x2) = [0 0]x0 + [0 1]x1 + [0 0]x2, [1] [c] = [0], [0] [b] = [1] orientation: [3 2] [4] [3 2] [3] active(f(b(),X,c())) = [0 0]X + [0] >= [0 0]X + [0] = mark(f(X,c(),X)) [1] [1] active(c()) = [0] >= [0] = mark(b()) [1 2] [3 2] [2 0] [1 2] [3 0] [2 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) [1 2] [3 3] [2 0] [1 2] [3 3] [2 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)) [1 2] [3 2] [2 0] [1 2] [3 2] [2 0] proper(f(X1,X2,X3)) = [0 0]X1 + [0 1]X2 + [0 0]X3 >= [0 0]X1 + [0 1]X2 + [0 0]X3 = f(proper(X1),proper(X2),proper(X3)) [0] [0] proper(b()) = [1] >= [1] = ok(b()) [1] [1] proper(c()) = [0] >= [0] = ok(c()) [1 2] [3 2] [2 0] [1 2] [3 2] [2 0] f(ok(X1),ok(X2),ok(X3)) = [0 0]X1 + [0 1]X2 + [0 0]X3 >= [0 0]X1 + [0 1]X2 + [0 0]X3 = ok(f(X1,X2,X3)) [2 2] [2] [2 0] [2] top(mark(X)) = [1 1]X + [0] >= [1 1]X + [0] = top(proper(X)) [2 0] [2] [2 0] [2] top(ok(X)) = [1 1]X + [0] >= [1 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=1 interpretation: [top](x0) = x0 + 2, [ok](x0) = x0, [proper](x0) = x0, [mark](x0) = x0 + 1, [active](x0) = x0,
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