# Pf there is no wff of length 2

• Apr 1st 2009, 03:55 PM
krystalballz
Pf there is no wff of length 2
How does one show that there are no wff's of length 2,3 or 6, but that any other positive length is possible? It seems obvious I just don't know how to express it as a proof.
• Apr 1st 2009, 07:31 PM
aliceinwonderland
Quote:

Originally Posted by krystalballz
How does one show that there are no wff's of length 2,3 or 6, but that any other positive length is possible? It seems obvious I just don't know how to express it as a proof.

Let L(S) be a length of an expression S.

Each sentence symbol x is a well-formed formula of L(x)=1.

For example,

For a wff a, $L(( \neg a)) = L(a) + 3$.
For wff a,b, $L((a \wedge b)) = L(a) + L(b) +3$.

Now, we have wff's of length 1, 4, and 5. You can check the remaining cases inductively and make sure that no wff's of length 2,3 or 6 is possible.