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TRS Standard pair #487067607
details
property
value
status
complete
benchmark
#3.24.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n074.star.cs.uiowa.edu
space
AG01
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
0.413544 seconds
cpu usage
0.642176
user time
0.482698
system time
0.159478
max virtual memory
96176.0
max residence set size
58800.0
stage attributes
key
value
starexec-result
YES
output
YES Problem: f(0()) -> s(0()) f(s(0())) -> s(0()) f(s(s(x))) -> f(f(s(x))) Proof: Matrix Interpretation Processor: dim=3 interpretation: [1 1 0] [f](x0) = [0 0 0]x0 [0 0 1] , [0] [0] = [1] [1], [1 0 0] [s](x0) = [0 0 0]x0 [0 0 1] orientation: [1] [0] f(0()) = [0] >= [0] = s(0()) [1] [1] [0] [0] f(s(0())) = [0] >= [0] = s(0()) [1] [1] [1 0 0] [1 0 0] f(s(s(x))) = [0 0 0]x >= [0 0 0]x = f(f(s(x))) [0 0 1] [0 0 1] problem: f(s(0())) -> s(0()) f(s(s(x))) -> f(f(s(x))) Matrix Interpretation Processor: dim=3 interpretation: [1 1 0] [0] [f](x0) = [0 0 1]x0 + [0] [0 0 0] [1], [0] [0] = [1] [0], [1 0 1] [0] [s](x0) = [0 0 0]x0 + [1] [0 1 0] [0] orientation: [1] [0] f(s(0())) = [1] >= [1] = s(0()) [1] [1] [1 1 1] [1] [1 1 1] [1] f(s(s(x))) = [0 0 0]x + [1] >= [0 0 0]x + [1] = f(f(s(x))) [0 0 0] [1] [0 0 0] [1] problem: f(s(s(x))) -> f(f(s(x))) Matrix Interpretation Processor: dim=3 interpretation: [1 0 1] [f](x0) = [0 0 1]x0 [0 0 0] , [1 0 0] [0] [s](x0) = [0 0 0]x0 + [1] [0 1 1] [0] orientation: [1 1 1] [1] [1 1 1] f(s(s(x))) = [0 1 1]x + [1] >= [0 0 0]x = f(f(s(x))) [0 0 0] [0] [0 0 0] problem: Qed
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