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TRS Standard pair #516966249
details
property
value
status
complete
benchmark
gen-1.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n128.star.cs.uiowa.edu
space
Secret_06_TRS
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
0.617382049561 seconds
cpu usage
1.136154672
max memory
1.91737856E8
stage attributes
key
value
output-size
2262
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_ttt2 /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES Problem: f(c(a(),z,x)) -> b(a(),z) b(x,b(z,y)) -> f(b(f(f(z)),c(x,z,y))) b(y,z) -> z Proof: Matrix Interpretation Processor: dim=1 interpretation: [c](x0, x1, x2) = 2x0 + 2x1 + x2 + 2, [b](x0, x1) = 4x0 + 2x1 + 2, [a] = 0, [f](x0) = x0 orientation: f(c(a(),z,x)) = x + 2z + 2 >= 2z + 2 = b(a(),z) b(x,b(z,y)) = 4x + 4y + 8z + 6 >= 4x + 2y + 8z + 6 = f(b(f(f(z)),c(x,z,y))) b(y,z) = 4y + 2z + 2 >= z = z problem: f(c(a(),z,x)) -> b(a(),z) b(x,b(z,y)) -> f(b(f(f(z)),c(x,z,y))) Matrix Interpretation Processor: dim=3 interpretation: [1 0 0] [1 1 1] [1 1 0] [1] [c](x0, x1, x2) = [0 0 0]x0 + [0 1 1]x1 + [0 1 1]x2 + [0] [0 0 0] [0 0 0] [0 0 0] [0], [1 0 1] [1 1 1] [0] [b](x0, x1) = [1 1 0]x0 + [0 1 1]x1 + [0] [1 1 1] [1 1 1] [1], [0] [a] = [0] [0], [1 0 0] [f](x0) = [0 1 0]x0 [1 0 0] orientation: [1 1 0] [1 1 1] [1] [1 1 1] [0] f(c(a(),z,x)) = [0 1 1]x + [0 1 1]z + [0] >= [0 1 1]z + [0] = b(a(),z) [1 1 0] [1 1 1] [1] [1 1 1] [1] [1 0 1] [2 3 3] [3 2 2] [1] [1 0 0] [1 2 1] [3 2 2] [1] b(x,b(z,y)) = [1 1 0]x + [1 2 2]y + [2 2 1]z + [1] >= [0 0 0]x + [0 1 1]y + [1 2 1]z + [0] = f(b(f(f(z)),c(x,z,y))) [1 1 1] [2 3 3] [3 2 2] [2] [1 0 0] [1 2 1] [3 2 2] [1] problem: b(x,b(z,y)) -> f(b(f(f(z)),c(x,z,y))) Matrix Interpretation Processor: dim=1 interpretation: [c](x0, x1, x2) = x0 + x1 + 2x2 + 2, [b](x0, x1) = 2x0 + 2x1 + 6, [f](x0) = x0 + 1 orientation: b(x,b(z,y)) = 2x + 4y + 4z + 18 >= 2x + 4y + 4z + 15 = f(b(f(f(z)),c(x,z,y))) problem: Qed
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