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SRS Standard pair #487091414
details
property
value
status
complete
benchmark
z121.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n141.star.cs.uiowa.edu
space
Zantema_04
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
2.241 seconds
cpu usage
7.76298
user time
6.20529
system time
1.55769
max virtual memory
5958700.0
max residence set size
68496.0
stage attributes
key
value
starexec-result
YES
output
YES Problem: f(f(x1)) -> b(b(b(x1))) a(f(x1)) -> f(a(a(x1))) b(b(x1)) -> c(c(a(c(x1)))) d(b(x1)) -> d(a(b(x1))) c(c(x1)) -> d(d(d(x1))) b(d(x1)) -> d(b(x1)) c(d(d(x1))) -> f(x1) Proof: Matrix Interpretation Processor: dim=1 interpretation: [d](x0) = x0 + 4, [b](x0) = x0 + 9, [c](x0) = x0 + 6, [f](x0) = x0 + 14, [a](x0) = x0 orientation: f(f(x1)) = x1 + 28 >= x1 + 27 = b(b(b(x1))) a(f(x1)) = x1 + 14 >= x1 + 14 = f(a(a(x1))) b(b(x1)) = x1 + 18 >= x1 + 18 = c(c(a(c(x1)))) d(b(x1)) = x1 + 13 >= x1 + 13 = d(a(b(x1))) c(c(x1)) = x1 + 12 >= x1 + 12 = d(d(d(x1))) b(d(x1)) = x1 + 13 >= x1 + 13 = d(b(x1)) c(d(d(x1))) = x1 + 14 >= x1 + 14 = f(x1) problem: a(f(x1)) -> f(a(a(x1))) b(b(x1)) -> c(c(a(c(x1)))) d(b(x1)) -> d(a(b(x1))) c(c(x1)) -> d(d(d(x1))) b(d(x1)) -> d(b(x1)) c(d(d(x1))) -> f(x1) Matrix Interpretation Processor: dim=1 interpretation: [d](x0) = x0 + 1, [b](x0) = 4x0 + 2, [c](x0) = 2x0 + 1, [f](x0) = 2x0 + 4, [a](x0) = x0 orientation: a(f(x1)) = 2x1 + 4 >= 2x1 + 4 = f(a(a(x1))) b(b(x1)) = 16x1 + 10 >= 8x1 + 7 = c(c(a(c(x1)))) d(b(x1)) = 4x1 + 3 >= 4x1 + 3 = d(a(b(x1))) c(c(x1)) = 4x1 + 3 >= x1 + 3 = d(d(d(x1))) b(d(x1)) = 4x1 + 6 >= 4x1 + 3 = d(b(x1)) c(d(d(x1))) = 2x1 + 5 >= 2x1 + 4 = f(x1) problem: a(f(x1)) -> f(a(a(x1))) d(b(x1)) -> d(a(b(x1))) c(c(x1)) -> d(d(d(x1))) String Reversal Processor: f(a(x1)) -> a(a(f(x1))) b(d(x1)) -> b(a(d(x1))) c(c(x1)) -> d(d(d(x1))) Bounds Processor: bound: 0 enrichment: match automaton: final states: {8,5,1} transitions: f50() -> 2* f0(2) -> 3* a0(6) -> 7* a0(4) -> 1* a0(3) -> 4* b0(7) -> 5* d0(2) -> 6* d0(6) -> 9* d0(9) -> 8* 1 -> 3* problem: Qed
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