Okay, so here is the deal. I had my first logic class today, I actually missed 2 of them since I wasn’t registered in the class. (I also didn’t have a working writing utility today…but um, whatever. I wrote my notes in blood. Mwa ha ha… …or, you know, with some guy’s pen.)…anyway, I’m going to attempt at explaining the lecture…. I ended up getting too interested and didn’t really take any notes. So, this may be sort of fragmented and may also…not make any sense.
Introduction
principle: if S is a set of well-formed formulas containing all sentence
symbols and close under all S formula-building functions, then S contains all
the wffs.
arbitrary wff. It is the last number of some construction sequence <e,
…,en.> where en will equal α.
Introduction
on natural integers to show every number of the sequence in S.
Base Case:
e,. e must be a sentence symbol, so it is in S.
Inductive
step: suppose that for all k<i, ek,eS.
Consider
ei. Either ei is a sentence symbol, so it is in S.
Or ei = Є¬
(ek) from k<i. ei = (¬ek)
Or ei = Є^
(ek,ej) for some ej, k<I ei = (ex^ej)
the inductive hypothesis that ek and ej are both S. S is closed under the
formula building functions. So ei is in S.
we can prove basic thing, such as an obviously basic statement as this: show
that every wffs has the same number of left and right parentheses. We call a
wff with this property, balanced.
Base case:
sentence symbols have no parentheses so they’re balanced.
Inductive
step: Suppose α and β are wffs. (This is also our inductive hypothesis.)
Let α have
n left and right parentheses and let β have m.
Case ¬: (¬α)
has n+1 left and right. So (¬β) is balanced.
Case ^: (α^
β) has n+m+1 left and n+m+1 right. So it is balanced.
Since this
is syntax, we can say (α |-| β)
Where |-|
is a binary connective.
So every
wff is balanced.
that, as I said is pure syntax. However, with that and from there we can ask
the thorny philosophical question ‘what is truth’?
assignment v for a set S of sentence symbol s to be a function. V:s –>
{T,F}
The
objective we want to achieve is a truth about compound statements based on
simple statements.
Define |S
to be the set of wffs built up from sentence symbols in S.
[Suppose
S={A1, A4, A28} |S = all wffs using
these sentence symbols
(A4 --> (A28^A4))]
want, is to extend v to |v such that |v:|S –> {T,F}
1) For any Ai Є S, |v (Ai)=v(Ai)
2) |v ((¬α)) = {T if |v(α) = F. F
otherwise.}
3) |v ((α^β)) = {T if |v (α) = T and |v
(β) = T. F otherwise}
4) |v ((αVβ)) = {T if |v (α) = T or |v
(β) = T (or both). F otherwise}
5) |v ((α –>β))
= {F if |v (α) = T and |v (β) = F. T otherwise}
6) |v ((α↔β)) = {T if |v (α) = |v (β).
F otherwise}
Now, having said all that, I have about 3 questions, so if you a) understand what I just typed out or b) think you could answer some questions on it well enough to be on the same level as a 3rd year philosophy of logic class - please, comment or message me. My prof isn’t taking appointments right now, and her office hours happen to be when I have a class. Thannnks.
