Re: Here is a hard one, for me...

On Sun, 22 May 2005, 1st Semester Logic Student wrote:

> Maybe I should also add the "hint."
>
Where's the problem statement.

> It says:
>
> "For shorthand, use the notion "o[P]" to stand for "the number of open
> brackets in P", and "b[P]" to stand for "the number of binary
> connectives in P". So you must prove that every sentence P has the
> following property: that o[P] = b[P]. First, show that all the sentence
> letters have this property (ie. show that all the sentence letters have
> the same number of open brackets as binary connectives). This is the
> "basis" of the proof. Second, show that each of the clauses in the
> definition of sentences of SL preserves this property (e.g. for the
> sixth clause, show that is both P and Q have the same number of open
> brackets as binary connectives, then (P tripplebar Q) has the same
> number of open brackets as binary connectives). This is the "inductive
> step" of the proof"
>
> That is the "hint" given with the original problem given in the first
> post. Hope that helps!
>
That's what I was outlining in my post. Let's see what details you can
put to it. Do take note, that only you know what the rules and system of
logic you are studying for there are many variations on the propositional
calculus both in construction of wff's, connectives, logical connectives
and even axioms. That is to say, as we're not mind readers, you need to
fill in these details as you proceed showing us your work.
.

Loading