Tool CaT
stdout:
MAYBE
Problem:
p(0()) -> 0()
p(s(x)) -> x
plus(x,0()) -> x
plus(0(),y) -> y
plus(s(x),y) -> s(plus(x,y))
plus(s(x),y) -> s(plus(p(s(x)),y))
plus(x,s(y)) -> s(plus(x,p(s(y))))
times(0(),y) -> 0()
times(s(0()),y) -> y
times(s(x),y) -> plus(y,times(x,y))
div(0(),y) -> 0()
div(x,y) -> quot(x,y,y)
quot(zero(y),s(y),z) -> 0()
quot(s(x),s(y),z) -> quot(x,y,z)
quot(x,0(),s(z)) -> s(div(x,s(z)))
div(div(x,y),z) -> div(x,times(zero(y),z))
eq(0(),0()) -> true()
eq(s(x),0()) -> false()
eq(0(),s(y)) -> false()
eq(s(x),s(y)) -> eq(x,y)
divides(y,x) -> eq(x,times(div(x,y),y))
prime(s(s(x))) -> pr(s(s(x)),s(x))
pr(x,s(0())) -> true()
pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
if(true(),x,y) -> false()
if(false(),x,y) -> pr(x,y)
zero(div(x,x)) -> x
zero(divides(x,x)) -> x
zero(times(x,x)) -> x
zero(quot(x,x,x)) -> x
zero(s(x)) -> if(eq(x,s(0())),plus(zero(0()),0()),s(plus(0(),zero(0()))))
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:
{ p(0()) -> 0()
, p(s(x)) -> x
, plus(x, 0()) -> x
, plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, y))
, plus(s(x), y) -> s(plus(p(s(x)), y))
, plus(x, s(y)) -> s(plus(x, p(s(y))))
, times(0(), y) -> 0()
, times(s(0()), y) -> y
, times(s(x), y) -> plus(y, times(x, y))
, div(0(), y) -> 0()
, div(x, y) -> quot(x, y, y)
, quot(zero(y), s(y), z) -> 0()
, quot(s(x), s(y), z) -> quot(x, y, z)
, quot(x, 0(), s(z)) -> s(div(x, s(z)))
, div(div(x, y), z) -> div(x, times(zero(y), z))
, eq(0(), 0()) -> true()
, eq(s(x), 0()) -> false()
, eq(0(), s(y)) -> false()
, eq(s(x), s(y)) -> eq(x, y)
, divides(y, x) -> eq(x, times(div(x, y), y))
, prime(s(s(x))) -> pr(s(s(x)), s(x))
, pr(x, s(0())) -> true()
, pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
, if(true(), x, y) -> false()
, if(false(), x, y) -> pr(x, y)
, zero(div(x, x)) -> x
, zero(divides(x, x)) -> x
, zero(times(x, x)) -> x
, zero(quot(x, x, x)) -> x
, zero(s(x)) ->
if(eq(x, s(0())), plus(zero(0()), 0()), s(plus(0(), zero(0()))))}
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:
{ p(0()) -> 0()
, p(s(x)) -> x
, plus(x, 0()) -> x
, plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, y))
, plus(s(x), y) -> s(plus(p(s(x)), y))
, plus(x, s(y)) -> s(plus(x, p(s(y))))
, times(0(), y) -> 0()
, times(s(0()), y) -> y
, times(s(x), y) -> plus(y, times(x, y))
, div(0(), y) -> 0()
, div(x, y) -> quot(x, y, y)
, quot(zero(y), s(y), z) -> 0()
, quot(s(x), s(y), z) -> quot(x, y, z)
, quot(x, 0(), s(z)) -> s(div(x, s(z)))
, div(div(x, y), z) -> div(x, times(zero(y), z))
, eq(0(), 0()) -> true()
, eq(s(x), 0()) -> false()
, eq(0(), s(y)) -> false()
, eq(s(x), s(y)) -> eq(x, y)
, divides(y, x) -> eq(x, times(div(x, y), y))
, prime(s(s(x))) -> pr(s(s(x)), s(x))
, pr(x, s(0())) -> true()
, pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
, if(true(), x, y) -> false()
, if(false(), x, y) -> pr(x, y)
, zero(div(x, x)) -> x
, zero(divides(x, x)) -> x
, zero(times(x, x)) -> x
, zero(quot(x, x, x)) -> x
, zero(s(x)) ->
if(eq(x, s(0())), plus(zero(0()), 0()), s(plus(0(), zero(0()))))}
Proof Output:
Computation stopped due to timeout after 60.0 seconds