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TRS Standard pair #516964684
details
property
value
status
complete
benchmark
koen.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n175.star.cs.uiowa.edu
space
Rubio_04
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
0.47739815712 seconds
cpu usage
0.662389571
max memory
1.35692288E8
stage attributes
key
value
output-size
1651
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: f(s(X),X) -> f(X,a(X)) f(X,c(X)) -> f(s(X),X) f(X,X) -> c(X) Proof: Matrix Interpretation Processor: dim=1 interpretation: [f](x0, x1) = 2x0 + 2x1 + 3, [c](x0) = 4x0, [s](x0) = 4x0, [a](x0) = 4x0 orientation: f(s(X),X) = 10X + 3 >= 10X + 3 = f(X,a(X)) f(X,c(X)) = 10X + 3 >= 10X + 3 = f(s(X),X) f(X,X) = 4X + 3 >= 4X = c(X) problem: f(s(X),X) -> f(X,a(X)) f(X,c(X)) -> f(s(X),X) Matrix Interpretation Processor: dim=1 interpretation: [f](x0, x1) = x0 + x1 + 4, [c](x0) = 7x0 + 1, [s](x0) = 7x0, [a](x0) = 7x0 orientation: f(s(X),X) = 8X + 4 >= 8X + 4 = f(X,a(X)) f(X,c(X)) = 8X + 5 >= 8X + 4 = f(s(X),X) problem: f(s(X),X) -> f(X,a(X)) Matrix Interpretation Processor: dim=3 interpretation: [1 1 0] [1 1 0] [0] [f](x0, x1) = [1 1 0]x0 + [1 1 0]x1 + [0] [1 0 1] [0 0 1] [1], [1 0 0] [1] [s](x0) = [1 0 0]x0 + [0] [1 0 0] [0], [1 0 0] [0] [a](x0) = [1 0 0]x0 + [0] [1 0 0] [1] orientation: [3 1 0] [1] [3 1 0] [0] f(s(X),X) = [3 1 0]X + [1] >= [3 1 0]X + [0] = f(X,a(X)) [2 0 1] [2] [2 0 1] [2] problem: Qed
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