views:

54

answers:

3

R(x) is a red block
B(x) is a blue block
T(x,y) block x is on top of block y

Question:
Write a formula asserting that if no red block is on top of a red block then no red block is on top of itself.

My answer: (Ax)(Ay)(R(x) and R(y) -> ~T(x,y))->(Ax)(R(x)-> ~T(x,x))
A = For all
~ = Not
-> = implies

A: 

Looks good to me.

David
+2  A: 

That is a plausible formulation, though not necessarily the most straight-forward translation of the sentence, which, to my mind, is (Ax)(Ay)(T(x,y) -> R(x) -> ~R(y)) -> ~(3x)(R(x) and T(x,x)). 3, here, being the existential quantifier (i.e. "there exists an").

ig0774
i m sorry, i havent tried yet, but do you think this formula is closed? in otherwords, has a tableau proof?
is this addressing my formulation or yours? my formulation was somewhat off-hand. I think they're equivalent and both closed, but your formulation looks as if it would have a shorter proof.
ig0774
cool. thanks man.
i'll add the caveat that it's been a few years since i've done a formal proof...
ig0774
A: 

(Ax)(Ay)(Az) { (((R(x) and R(y)) -> ~T(x,y)) -> ~T(z,z) }

RBarryYoung