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TRS Standard pair #487074042
details
property
value
status
complete
benchmark
filliatre.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n192.star.cs.uiowa.edu
space
CiME_04
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
0.795406 seconds
cpu usage
2.08904
user time
1.64545
system time
0.44359
max virtual memory
1156032.0
max residence set size
64072.0
stage attributes
key
value
starexec-result
YES
output
YES Problem: g(A()) -> A() g(B()) -> A() g(B()) -> B() g(C()) -> A() g(C()) -> B() g(C()) -> C() foldf(x,nil()) -> x foldf(x,cons(y,z)) -> f(foldf(x,z),y) f(t,x) -> f'(t,g(x)) f'(triple(a,b,c),C()) -> triple(a,b,cons(C(),c)) f'(triple(a,b,c),B()) -> f(triple(a,b,c),A()) f'(triple(a,b,c),A()) -> f''(foldf(triple(cons(A(),a),nil(),c),b)) f''(triple(a,b,c)) -> foldf(triple(a,b,nil()),c) Proof: Matrix Interpretation Processor: dim=1 interpretation: [nil] = 0, [f'](x0, x1) = x0 + 2x1, [g](x0) = x0, [C] = 1, [foldf](x0, x1) = x0 + 2x1, [f''](x0) = x0 + 1, [A] = 1, [f](x0, x1) = x0 + 2x1, [triple](x0, x1, x2) = x0 + 4x1 + 2x2, [cons](x0, x1) = x0 + x1, [B] = 1 orientation: g(A()) = 1 >= 1 = A() g(B()) = 1 >= 1 = A() g(B()) = 1 >= 1 = B() g(C()) = 1 >= 1 = A() g(C()) = 1 >= 1 = B() g(C()) = 1 >= 1 = C() foldf(x,nil()) = x >= x = x foldf(x,cons(y,z)) = x + 2y + 2z >= x + 2y + 2z = f(foldf(x,z),y) f(t,x) = t + 2x >= t + 2x = f'(t,g(x)) f'(triple(a,b,c),C()) = a + 4b + 2c + 2 >= a + 4b + 2c + 2 = triple(a,b,cons(C(),c)) f'(triple(a,b,c),B()) = a + 4b + 2c + 2 >= a + 4b + 2c + 2 = f(triple(a,b,c),A()) f'(triple(a,b,c),A()) = a + 4b + 2c + 2 >= a + 2b + 2c + 2 = f''(foldf(triple(cons(A(),a),nil(),c),b)) f''(triple(a,b,c)) = a + 4b + 2c + 1 >= a + 4b + 2c = foldf(triple(a,b,nil()),c) problem: g(A()) -> A() g(B()) -> A() g(B()) -> B() g(C()) -> A() g(C()) -> B() g(C()) -> C() foldf(x,nil()) -> x foldf(x,cons(y,z)) -> f(foldf(x,z),y) f(t,x) -> f'(t,g(x)) f'(triple(a,b,c),C()) -> triple(a,b,cons(C(),c)) f'(triple(a,b,c),B()) -> f(triple(a,b,c),A()) f'(triple(a,b,c),A()) -> f''(foldf(triple(cons(A(),a),nil(),c),b)) Matrix Interpretation Processor: dim=1 interpretation: [nil] = 4, [f'](x0, x1) = x0 + 5x1, [g](x0) = x0, [C] = 6, [foldf](x0, x1) = x0 + 5x1, [f''](x0) = x0, [A] = 5, [f](x0, x1) = x0 + 5x1,
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