Propositional logic, predicate logic and equations

Question

I have $$ 6y=x^2 \quad ∧\quad (y=2 \quad ∨\quad y=-8)$$$$ \iff (6y=x^2 \quad ∧ \quad y=2 ) \quad ∨\quad (6y=x^2 \quad ∧\quad y=-8). \tag1$$ Is it correct to say that step (1) is valid because $$ A ∧(B ∨ C) ⇔ (A∧B)∨ (A∧C)\tag2$$ is a tautology?

But does (2) being a tautology really legitimize what I did in (1)? After all (2) belongs to propositional logic while (1) belongs to predicate logic.

Here is my attempt to answer my question:

Yes, I would think that it is enough to show that (2) is a tautology to legitimize what I did in (1). The reason is that the truth table of (2) contains ALL the possible combinations of logical values (“true” or ” false”) of $A$, $B$ and $C$. Even though $6y=x^2$ , $y=2$ and $y=-8$ are predicates, they turn into a true or false sentence after I plugged in certain numbers. For example, when I plug in $y=2$ ,$x=\sqrt{12}$ into (1): this $6y=x^2$ and this $y=2$ becomes true, while this $y=-8$ becomes false. But that’s a situation I can find in the truth table of (2) as the row where $$ A=1 \ \ \ \ B=1 \ \ \ \ C=0 \ \ . $$ For everything that can happen in (1) I can find a corresponding row from the truth table of (2). That’s why the truth table of (2) “covers” everything that can happen in (1). And that’s why the tautology in (2) legitimizes the step in (1).

Answer 1

$$ 6y=x^2 \quad ∧\quad (y=2 \quad ∨\quad y=-8)$$$$ \iff (6y=x^2 \quad ∧ \quad y=2 ) \quad ∨\quad (6y=x^2 \quad ∧\quad y=-8). \tag1$$ does $$A ∧(B ∨ C) ⇔ (A∧B)∨ (A∧C)\tag2$$ being a tautology really legitimize what I did in (1)? After all (2) belongs to propositional logic while (1) belongs to predicate logic.

The proposition $Q$ is the truth-functional form of the proposition $\big(\exists x\;(x>0 \land x<10)\big).$

The proposition $A$ is the truth-functional form of the predicate $Axy.$

By the substitutions \begin{align}A&:=(Axy:=)&&6y=x^2,\\B&:=(By:=)&&y=2,\\C&:=(Cy:=)&&y=-8,\end{align} the proposition $(2)$ is the truth-functional form of the predicate $(1).$

Thus, since $(2)$ is a tautology, $(1)$ is also a tautology. Notice that this argument was made in point #2 of our previous discussion. We say that the two sides of $(1)$ are tautologically equivalent to each other.

Naturally, $(1)$'s formalisation $$Axy ∧(By ∨ Cy) ⇔ (Axy∧By)∨ (Axy∧Cy)\tag3$$ is also a tautology.

Predicate logic is an extension of, not complementary to, propositional logic.

Pertinently, your above analysis is applicable only because $(1)$ involves no quantification, because it is in the midst of dealing with arbitrary $x$ and $y.$

Addendum

Could you pls explain to me what you mean exactly by "truth-functional form"?

Sure; let me know if this algorithm is unclear. To determine the truth-functional form of a predicate-logic formula:

1. working from left-to-right, underline the entire scope of each quantifier, including the quantifier itself;
2. underline each remaining atomic formula;
3. assign a symbol—usually a capital letter, representing an atomic propositional-logic proposition—to every underlined formula, assigning the same symbol to two formulae if and only if they are identical.