I am attempting to read through the Alfred Tarski's works on logic. Unfortunately I have not gotten very far before hitting a wall; I am hoping someone can shed light on something that seems elementary but I have been unable to wrap my head around. In his paper On the Primitive Term of the Logistic, he claims all logical operations can be defined in terms of quantification, the equivalence relation and conjunction. He gives an example of how to define negation using only universal quantification and equivalences,
$\forall p: \neg p \equiv (p \equiv \forall q: q) $
He uses it in the proof of Theorem 6 in the paper as if it were a well known identity that does not need further explanation, but I have been unable to find anything on it in any of my searches. So, I have been trying to understand how to interpret this identity on my own without much success. The term $ \forall q: q $ is causing me the most confusion. It seems to be saying, "for all propositions q, q is true". Under this interpretation, I am not sure how to understand the alleged identity. Let me explain my reasoning so far:
$ \forall q: q $ is a proposition without any free variables. Its truth depends on the values assigned to the class of propositions considered. When the class of all propositions contains nothing but values of true, the universal proposition is true. If at least one proposition is false, then the universal proposition is false.
To evaluate the equivalence $ \neg p \equiv ( p \equiv \forall q: q) $, I am thinking of it in terms of value assignments of the component propositions,
| p | $\forall q: q$ |
|---|---|
| T | T |
| T | F |
| F | T |
| F | F |
Now, to my mind, the third row of the table is an impossibility, since if $\forall q: q$ is true, then $p$ can't be assigned a value of false, otherwise the universal would be false. So, I've struck that from the table,
| p | $\forall q: q$ |
|---|---|
| T | T |
| T | F |
| F | F |
In the first and third case, the nested equivalence is true, so the outer equivalence would be false. In the second case, the nested equivalence is false, so the outer equivalence would be true (since $p$ is true at that assignment).
At this point, I feel like I have made a mistake along the way somewhere, but I can't pinpoint what the flaw in my reasoning is.
We can define negation with the conditional: $\to$, and falsum: $\bot$, a propositional constant that is always False:
Thus, using quantified propositional logic, we can use $\forall q.q$ in place of falsum, because $\forall q.q$ is always False.
If you want to use the bi-conditional, we have to check the truth table with formulas: $p$ and $p \equiv \forall q.q$.
When $p$ is True the second formula is False (because $\text T \to \text F$ is $\text F$) and when $p$ is False the second formula is True (because $\text F \to \text F$ is $\text T$).
In a nutshell, you have to delete the first line in the truth table above (second one).