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SRS Standard pair #487089380
details
property
value
status
complete
benchmark
aprove2.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n138.star.cs.uiowa.edu
space
Secret_05_SRS
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
3.53695 seconds
cpu usage
12.83
user time
10.7974
system time
2.03262
max virtual memory
6034880.0
max residence set size
71092.0
stage attributes
key
value
starexec-result
YES
output
YES Problem: g(h(x1)) -> g(f(s(x1))) f(s(s(s(x1)))) -> h(f(s(h(x1)))) f(h(x1)) -> h(f(s(h(x1)))) h(x1) -> x1 f(f(s(s(x1)))) -> s(s(s(f(f(x1))))) b(a(x1)) -> a(b(x1)) a(a(a(x1))) -> b(a(a(b(x1)))) b(b(b(b(x1)))) -> a(x1) Proof: String Reversal Processor: h(g(x1)) -> s(f(g(x1))) s(s(s(f(x1)))) -> h(s(f(h(x1)))) h(f(x1)) -> h(s(f(h(x1)))) h(x1) -> x1 s(s(f(f(x1)))) -> f(f(s(s(s(x1))))) a(b(x1)) -> b(a(x1)) a(a(a(x1))) -> b(a(a(b(x1)))) b(b(b(b(x1)))) -> a(x1) Matrix Interpretation Processor: dim=1 interpretation: [a](x0) = x0 + 2, [g](x0) = x0 + 8, [f](x0) = x0, [b](x0) = x0 + 1, [h](x0) = x0, [s](x0) = x0 orientation: h(g(x1)) = x1 + 8 >= x1 + 8 = s(f(g(x1))) s(s(s(f(x1)))) = x1 >= x1 = h(s(f(h(x1)))) h(f(x1)) = x1 >= x1 = h(s(f(h(x1)))) h(x1) = x1 >= x1 = x1 s(s(f(f(x1)))) = x1 >= x1 = f(f(s(s(s(x1))))) a(b(x1)) = x1 + 3 >= x1 + 3 = b(a(x1)) a(a(a(x1))) = x1 + 6 >= x1 + 6 = b(a(a(b(x1)))) b(b(b(b(x1)))) = x1 + 4 >= x1 + 2 = a(x1) problem: h(g(x1)) -> s(f(g(x1))) s(s(s(f(x1)))) -> h(s(f(h(x1)))) h(f(x1)) -> h(s(f(h(x1)))) h(x1) -> x1 s(s(f(f(x1)))) -> f(f(s(s(s(x1))))) a(b(x1)) -> b(a(x1)) a(a(a(x1))) -> b(a(a(b(x1)))) Matrix Interpretation Processor: dim=1 interpretation: [a](x0) = x0 + 4, [g](x0) = 2x0 + 8, [f](x0) = x0, [b](x0) = x0, [h](x0) = x0, [s](x0) = x0 orientation: h(g(x1)) = 2x1 + 8 >= 2x1 + 8 = s(f(g(x1))) s(s(s(f(x1)))) = x1 >= x1 = h(s(f(h(x1)))) h(f(x1)) = x1 >= x1 = h(s(f(h(x1)))) h(x1) = x1 >= x1 = x1 s(s(f(f(x1)))) = x1 >= x1 = f(f(s(s(s(x1))))) a(b(x1)) = x1 + 4 >= x1 + 4 = b(a(x1)) a(a(a(x1))) = x1 + 12 >= x1 + 8 = b(a(a(b(x1)))) problem: h(g(x1)) -> s(f(g(x1))) s(s(s(f(x1)))) -> h(s(f(h(x1)))) h(f(x1)) -> h(s(f(h(x1)))) h(x1) -> x1 s(s(f(f(x1)))) -> f(f(s(s(s(x1))))) a(b(x1)) -> b(a(x1)) String Reversal Processor: g(h(x1)) -> g(f(s(x1))) f(s(s(s(x1)))) -> h(f(s(h(x1)))) f(h(x1)) -> h(f(s(h(x1)))) h(x1) -> x1
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