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TRS Standard pair #516967804
details
property
value
status
complete
benchmark
filliatre2.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n068.star.cs.uiowa.edu
space
CiME_04
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
1.07554197311 seconds
cpu usage
2.846373703
max memory
2.55033344E8
stage attributes
key
value
output-size
12466
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: g(A()) -> A() g(B()) -> A() g(B()) -> B() g(C()) -> A() g(C()) -> B() g(C()) -> C() foldB(t,0()) -> t foldB(t,s(n)) -> f(foldB(t,n),B()) foldC(t,0()) -> t foldC(t,s(n)) -> f(foldC(t,n),C()) f(t,x) -> f'(t,g(x)) f'(triple(a,b,c),C()) -> triple(a,b,s(c)) f'(triple(a,b,c),B()) -> f(triple(a,b,c),A()) f'(triple(a,b,c),A()) -> f''(foldB(triple(s(a),0(),c),b)) f''(triple(a,b,c)) -> foldC(triple(a,b,0()),c) Proof: Matrix Interpretation Processor: dim=1 interpretation: [0] = 0, [foldC](x0, x1) = x0 + 2x1, [g](x0) = x0, [C] = 4, [foldB](x0, x1) = x0 + 2x1, [triple](x0, x1, x2) = x0 + 2x1 + 2x2 + 5, [A] = 2, [f](x0, x1) = x0 + 2x1, [f''](x0) = x0, [f'](x0, x1) = x0 + 2x1, [s](x0) = x0 + 4, [B] = 4 orientation: g(A()) = 2 >= 2 = A() g(B()) = 4 >= 2 = A() g(B()) = 4 >= 4 = B() g(C()) = 4 >= 2 = A() g(C()) = 4 >= 4 = B() g(C()) = 4 >= 4 = C() foldB(t,0()) = t >= t = t foldB(t,s(n)) = 2n + t + 8 >= 2n + t + 8 = f(foldB(t,n),B()) foldC(t,0()) = t >= t = t foldC(t,s(n)) = 2n + t + 8 >= 2n + t + 8 = f(foldC(t,n),C()) f(t,x) = t + 2x >= t + 2x = f'(t,g(x)) f'(triple(a,b,c),C()) = a + 2b + 2c + 13 >= a + 2b + 2c + 13 = triple(a,b,s(c)) f'(triple(a,b,c),B()) = a + 2b + 2c + 13 >= a + 2b + 2c + 9 = f(triple(a,b,c),A()) f'(triple(a,b,c),A()) = a + 2b + 2c + 9 >= a + 2b + 2c + 9 = f''(foldB(triple(s(a),0(),c),b)) f''(triple(a,b,c)) = a + 2b + 2c + 5 >= a + 2b + 2c + 5 = foldC(triple(a,b,0()),c) problem: g(A()) -> A() g(B()) -> B() g(C()) -> B() g(C()) -> C() foldB(t,0()) -> t foldB(t,s(n)) -> f(foldB(t,n),B()) foldC(t,0()) -> t foldC(t,s(n)) -> f(foldC(t,n),C()) f(t,x) -> f'(t,g(x)) f'(triple(a,b,c),C()) -> triple(a,b,s(c)) f'(triple(a,b,c),A()) -> f''(foldB(triple(s(a),0(),c),b)) f''(triple(a,b,c)) -> foldC(triple(a,b,0()),c) Matrix Interpretation Processor: dim=3 interpretation: [0] [0] = [0]
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