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.

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- Apr 1st 2009, 03:55 PMkrystalballzPf 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 PMaliceinwonderland

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, $\displaystyle L(( \neg a)) = L(a) + 3$.

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

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.