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SRS Standard pair #487517607
details
property
value
status
complete
benchmark
torpa4.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n018.star.cs.uiowa.edu
space
Secret_05_SRS
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
2.59954810143 seconds
cpu usage
8.910503374
max memory
1.119789056E9
stage attributes
key
value
output-size
2645
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: a(b(c(a(x1)))) -> b(a(c(b(a(b(x1)))))) a(d(x1)) -> c(x1) a(f(f(x1))) -> g(x1) b(g(x1)) -> g(b(x1)) c(x1) -> f(f(x1)) c(a(c(x1))) -> b(c(a(b(c(x1))))) c(d(x1)) -> a(a(x1)) g(x1) -> c(a(x1)) g(x1) -> d(d(d(d(x1)))) Proof: Matrix Interpretation Processor: dim=1 interpretation: [f](x0) = x0 + 4, [c](x0) = x0 + 8, [d](x0) = x0 + 3, [g](x0) = x0 + 13, [a](x0) = x0 + 5, [b](x0) = x0 orientation: a(b(c(a(x1)))) = x1 + 18 >= x1 + 18 = b(a(c(b(a(b(x1)))))) a(d(x1)) = x1 + 8 >= x1 + 8 = c(x1) a(f(f(x1))) = x1 + 13 >= x1 + 13 = g(x1) b(g(x1)) = x1 + 13 >= x1 + 13 = g(b(x1)) c(x1) = x1 + 8 >= x1 + 8 = f(f(x1)) c(a(c(x1))) = x1 + 21 >= x1 + 21 = b(c(a(b(c(x1))))) c(d(x1)) = x1 + 11 >= x1 + 10 = a(a(x1)) g(x1) = x1 + 13 >= x1 + 13 = c(a(x1)) g(x1) = x1 + 13 >= x1 + 12 = d(d(d(d(x1)))) problem: a(b(c(a(x1)))) -> b(a(c(b(a(b(x1)))))) a(d(x1)) -> c(x1) a(f(f(x1))) -> g(x1) b(g(x1)) -> g(b(x1)) c(x1) -> f(f(x1)) c(a(c(x1))) -> b(c(a(b(c(x1))))) g(x1) -> c(a(x1)) Matrix Interpretation Processor: dim=1 interpretation: [f](x0) = x0 + 1, [c](x0) = x0 + 4, [d](x0) = 4x0 + 4, [g](x0) = 4x0 + 8, [a](x0) = 4x0 + 1, [b](x0) = x0 orientation: a(b(c(a(x1)))) = 16x1 + 21 >= 16x1 + 21 = b(a(c(b(a(b(x1)))))) a(d(x1)) = 16x1 + 17 >= x1 + 4 = c(x1) a(f(f(x1))) = 4x1 + 9 >= 4x1 + 8 = g(x1) b(g(x1)) = 4x1 + 8 >= 4x1 + 8 = g(b(x1)) c(x1) = x1 + 4 >= x1 + 2 = f(f(x1)) c(a(c(x1))) = 4x1 + 21 >= 4x1 + 21 = b(c(a(b(c(x1))))) g(x1) = 4x1 + 8 >= 4x1 + 5 = c(a(x1)) problem: a(b(c(a(x1)))) -> b(a(c(b(a(b(x1)))))) b(g(x1)) -> g(b(x1)) c(a(c(x1))) -> b(c(a(b(c(x1))))) String Reversal Processor: a(c(b(a(x1)))) -> b(a(b(c(a(b(x1)))))) g(b(x1)) -> b(g(x1)) c(a(c(x1))) -> c(b(a(c(b(x1))))) Bounds Processor: bound: 0 enrichment: match automaton:
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