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SRS Stand 10685 pair #381723037
details
property
value
status
complete
benchmark
size-12-alpha-3-num-552.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n103.star.cs.uiowa.edu
space
Waldmann_07_size12
run statistics
property
value
solver
ttt2-1.17+nonreach
configuration
ttt2-1.17+nonreach
runtime (wallclock)
21.7182180882 seconds
cpu usage
85.552688223
max memory
9.24459008E8
stage attributes
key
value
output-size
2353
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES Problem: a(b(x1)) -> x1 a(c(x1)) -> b(c(c(a(a(b(x1)))))) b(c(x1)) -> x1 Proof: String Reversal Processor: b(a(x1)) -> x1 c(a(x1)) -> b(a(a(c(c(b(x1)))))) c(b(x1)) -> x1 DP Processor: DPs: c#(a(x1)) -> b#(x1) c#(a(x1)) -> c#(b(x1)) c#(a(x1)) -> c#(c(b(x1))) c#(a(x1)) -> b#(a(a(c(c(b(x1)))))) TRS: b(a(x1)) -> x1 c(a(x1)) -> b(a(a(c(c(b(x1)))))) c(b(x1)) -> x1 TDG Processor: DPs: c#(a(x1)) -> b#(x1) c#(a(x1)) -> c#(b(x1)) c#(a(x1)) -> c#(c(b(x1))) c#(a(x1)) -> b#(a(a(c(c(b(x1)))))) TRS: b(a(x1)) -> x1 c(a(x1)) -> b(a(a(c(c(b(x1)))))) c(b(x1)) -> x1 graph: c#(a(x1)) -> c#(c(b(x1))) -> c#(a(x1)) -> b#(a(a(c(c(b(x1)))))) c#(a(x1)) -> c#(c(b(x1))) -> c#(a(x1)) -> c#(c(b(x1))) c#(a(x1)) -> c#(c(b(x1))) -> c#(a(x1)) -> c#(b(x1)) c#(a(x1)) -> c#(c(b(x1))) -> c#(a(x1)) -> b#(x1) c#(a(x1)) -> c#(b(x1)) -> c#(a(x1)) -> b#(a(a(c(c(b(x1)))))) c#(a(x1)) -> c#(b(x1)) -> c#(a(x1)) -> c#(c(b(x1))) c#(a(x1)) -> c#(b(x1)) -> c#(a(x1)) -> c#(b(x1)) c#(a(x1)) -> c#(b(x1)) -> c#(a(x1)) -> b#(x1) SCC Processor: #sccs: 1 #rules: 2 #arcs: 8/16 DPs: c#(a(x1)) -> c#(c(b(x1))) c#(a(x1)) -> c#(b(x1)) TRS: b(a(x1)) -> x1 c(a(x1)) -> b(a(a(c(c(b(x1)))))) c(b(x1)) -> x1 Arctic Interpretation Processor: dimension: 1 usable rules: b(a(x1)) -> x1 c(a(x1)) -> b(a(a(c(c(b(x1)))))) c(b(x1)) -> x1 interpretation: [c#](x0) = x0 + 8, [c](x0) = 1x0 + 0, [a](x0) = 1x0 + 10, [b](x0) = -1x0 + 3 orientation: c#(a(x1)) = 1x1 + 10 >= x1 + 8 = c#(c(b(x1))) c#(a(x1)) = 1x1 + 10 >= -1x1 + 8 = c#(b(x1)) b(a(x1)) = x1 + 9 >= x1 = x1 c(a(x1)) = 2x1 + 11 >= 2x1 + 10 = b(a(a(c(c(b(x1)))))) c(b(x1)) = x1 + 4 >= x1 = x1 problem: DPs: TRS: b(a(x1)) -> x1 c(a(x1)) -> b(a(a(c(c(b(x1)))))) c(b(x1)) -> x1 Qed
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