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	n104.star.cs.uiowa.edu
space	CiME_04

run statistics

property	value
solver	ttt2-1.17+nonreach
configuration	ttt2-1.17+nonreach
runtime (wallclock)	1.13996219635 seconds
cpu usage	3.189774465
max memory	2.73412096E8

stage attributes

key	value
output-size	14193
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: 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: [f''](x0) = x0, [triple](x0, x1, x2) = x0 + 2x1 + x2, [f'](x0, x1) = x0 + x1 + 2, [foldC](x0, x1) = x0 + x1, [f](x0, x1) = x0 + 4x1 + 2, [s](x0) = x0 + 2, [foldB](x0, x1) = x0 + 2x1, [0] = 0, [C] = 0, [B] = 0, [g](x0) = 4x0, [A] = 0 orientation: g(A()) = 0 >= 0 = A() g(B()) = 0 >= 0 = A() g(B()) = 0 >= 0 = B() g(C()) = 0 >= 0 = A() g(C()) = 0 >= 0 = B() g(C()) = 0 >= 0 = C() foldB(t,0()) = t >= t = t foldB(t,s(n)) = 2n + t + 4 >= 2n + t + 2 = f(foldB(t,n),B()) foldC(t,0()) = t >= t = t foldC(t,s(n)) = n + t + 2 >= n + t + 2 = f(foldC(t,n),C()) f(t,x) = t + 4x + 2 >= t + 4x + 2 = f'(t,g(x)) f'(triple(a,b,c),C()) = a + 2b + c + 2 >= a + 2b + c + 2 = triple(a,b,s(c)) f'(triple(a,b,c),B()) = a + 2b + c + 2 >= a + 2b + c + 2 = f(triple(a,b,c),A()) f'(triple(a,b,c),A()) = a + 2b + c + 2 >= a + 2b + c + 2 = f''(foldB(triple(s(a),0(),c),b)) f''(triple(a,b,c)) = a + 2b + c >= a + 2b + c = foldC(triple(a,b,0()),c) problem: g(A()) -> A() g(B()) -> A() g(B()) -> B() g(C()) -> A() g(C()) -> B() g(C()) -> C() foldB(t,0()) -> t 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) Matrix Interpretation Processor: dim=3 interpretation: popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to TRS Stand 20472