Tool CaT
stdout:
MAYBE
Problem:
le(0(),y) -> true()
le(s(x),0()) -> false()
le(s(x),s(y)) -> le(x,y)
minus(x,0()) -> x
minus(s(x),s(y)) -> minus(x,y)
mod(0(),y) -> 0()
mod(s(x),0()) -> 0()
mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
if_mod(true(),x,y) -> mod(minus(x,y),y)
if_mod(false(),s(x),s(y)) -> s(x)
Proof:
OpenTool IRC1
stdout:
MAYBE
Tool IRC2
stdout:
TIMEOUT
'Fastest (timeout of 60.0 seconds)'
-----------------------------------
Answer: TIMEOUT
Input Problem: innermost runtime-complexity with respect to
Rules:
{ le(0(), y) -> true()
, le(s(x), 0()) -> false()
, le(s(x), s(y)) -> le(x, y)
, minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, mod(0(), y) -> 0()
, mod(s(x), 0()) -> 0()
, mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
, if_mod(true(), x, y) -> mod(minus(x, y), y)
, if_mod(false(), s(x), s(y)) -> s(x)}
Proof Output:
Computation stopped due to timeout after 60.0 secondsTool RC1
stdout:
MAYBE
Tool RC2
stdout:
TIMEOUT
'Fastest (timeout of 60.0 seconds)'
-----------------------------------
Answer: TIMEOUT
Input Problem: runtime-complexity with respect to
Rules:
{ le(0(), y) -> true()
, le(s(x), 0()) -> false()
, le(s(x), s(y)) -> le(x, y)
, minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, mod(0(), y) -> 0()
, mod(s(x), 0()) -> 0()
, mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
, if_mod(true(), x, y) -> mod(minus(x, y), y)
, if_mod(false(), s(x), s(y)) -> s(x)}
Proof Output:
Computation stopped due to timeout after 60.0 seconds