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SRS Stand 10685 pair #381718493
details
property
value
status
complete
benchmark
z104.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n093.star.cs.uiowa.edu
space
Zantema_04
run statistics
property
value
solver
ttt2-1.17+nonreach
configuration
ttt2-1.17+nonreach
runtime (wallclock)
1.92844605446 seconds
cpu usage
6.348821578
max memory
8.29329408E8
stage attributes
key
value
output-size
1941
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES Problem: c(c(c(a(x1)))) -> d(d(x1)) d(b(x1)) -> c(c(x1)) c(x1) -> a(a(a(a(x1)))) d(x1) -> b(b(b(b(x1)))) b(d(x1)) -> c(c(x1)) a(c(c(c(x1)))) -> d(d(x1)) Proof: Matrix Interpretation Processor: dim=1 interpretation: [b](x0) = x0 + 2, [d](x0) = x0 + 8, [c](x0) = x0 + 5, [a](x0) = x0 + 1 orientation: c(c(c(a(x1)))) = x1 + 16 >= x1 + 16 = d(d(x1)) d(b(x1)) = x1 + 10 >= x1 + 10 = c(c(x1)) c(x1) = x1 + 5 >= x1 + 4 = a(a(a(a(x1)))) d(x1) = x1 + 8 >= x1 + 8 = b(b(b(b(x1)))) b(d(x1)) = x1 + 10 >= x1 + 10 = c(c(x1)) a(c(c(c(x1)))) = x1 + 16 >= x1 + 16 = d(d(x1)) problem: c(c(c(a(x1)))) -> d(d(x1)) d(b(x1)) -> c(c(x1)) d(x1) -> b(b(b(b(x1)))) b(d(x1)) -> c(c(x1)) a(c(c(c(x1)))) -> d(d(x1)) Matrix Interpretation Processor: dim=1 interpretation: [b](x0) = x0, [d](x0) = 4x0, [c](x0) = 2x0, [a](x0) = 2x0 + 2 orientation: c(c(c(a(x1)))) = 16x1 + 16 >= 16x1 = d(d(x1)) d(b(x1)) = 4x1 >= 4x1 = c(c(x1)) d(x1) = 4x1 >= x1 = b(b(b(b(x1)))) b(d(x1)) = 4x1 >= 4x1 = c(c(x1)) a(c(c(c(x1)))) = 16x1 + 2 >= 16x1 = d(d(x1)) problem: d(b(x1)) -> c(c(x1)) d(x1) -> b(b(b(b(x1)))) b(d(x1)) -> c(c(x1)) String Reversal Processor: b(d(x1)) -> c(c(x1)) d(x1) -> b(b(b(b(x1)))) d(b(x1)) -> c(c(x1)) Bounds Processor: bound: 0 enrichment: match automaton: final states: {4,1} transitions: c0(2) -> 3* c0(3) -> 1* b0(5) -> 6* b0(7) -> 4* b0(2) -> 5* b0(6) -> 7* f40() -> 2* 1 -> 5* problem: Qed
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