Using logical equivalences without propositions?

Question

Discrete Mathematics, 8th edition, Rosen:

1. The set identities can be proved directly from the corresponding logical equivalences.
2. Compound propositions that have the same truth values in all possible cases are called logical equivalent.

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I want to use the laws of logical equivalences to prove set identities, for example, the distributive law $A\cup(B\cap C)=(A\cup B)\cap (A\cup C):$

$x\in A\cup(B\cap C)$

$\Leftrightarrow x\in A$ or ($x\in B$ and $x\in C$)

$\Leftrightarrow$ ($x\in A$ or $x\in B$) and ($x\in A$ or $x\in C$)

$\Leftrightarrow$ ......

To use the logical equivalence $p\vee(q\wedge r) \equiv (p\vee q)\wedge(p\vee r),$ $x\in A$, $x\in B$, $x\in C$ need to be propositions. I know that a proposition is any expression that is either true or false. Is '$x\in A$' True or False? Isn't $x\in A$ a propositional function? Yet I think that '$x\in A\cup(B\cap C)$' has to be a proposition, otherwise the proof becomes meaningless.

In the proof, is '$x\in A$' a proposition? If YES, why? If NO, how to use logical equivalences, which applies between propositions?

Answer 1

$x\in A\cup(B\cap C)$

$\Leftrightarrow x\in A$ or ($x\in B$ and $x\in C$)

$\Leftrightarrow$ ($x\in A$ or $x\in B$) and ($x\in A$ or $x\in C$)

$\Leftrightarrow$ ......

To use the logical equivalence $p\vee(q\wedge r) \equiv (p\vee q)\wedge(p\vee r),$ $x\in A$, $x\in B$, $x\in C$ need to be propositions.

Actually, this is a logical equivalence: $$P(x) ∨ (Q(x) ∧ R(x)) \quad\equiv\quad (P(x) ∨ Q(x)) ∧ (P(x) ∨ R(x)).$$

Isn't $x\in A$ a propositional function?

To be more explicit, you could change the proof's first line from

• $x\in A\cup(B\cap C)$

to

• $x$ is an arbitrary element of $A\cup(B\cap C),$

and similarly for its final line. As such $(x$ being an arbitrary constant rather than a free variable), $\text‘x\in A\text’$ is as legitimately a proposition—rather than a propositional function—as $\text‘e+\pi=13\text’$ is! Additionally, consider replacing the symbol $x$ with maybe $k$ so that it looks more like a parameter than a free variable.

I know that a proposition is any expression that is either true or false. Is '$x\in A$' True or False?

When the interpretation is not fixed, a proposition needn't generally have a definite truth value: for example, the proposition ‘The square of every number belongs to $[0,\infty)\text’$ is true in $\mathbb R$ and false in $\mathbb C.$ A better definition is:

• A proposition is a string of symbols that is well-formed according to the logic's syntactic rules and where no variable is free.

In any case, in the above proof we are asserting not that those propositions are true, merely that within each context they have a common truth value.

P.S. To be an arbitrary element of $A$ is to be a representative element of $A,$ without any restriction being imposed.

P.P.S. In the above proof, that sequence of logical equivalences is summarised by declaring that $$\forall x\,\big(x\in A\cup(B\cap C)\iff x\in(A\cup B)\cap (A\cup C)\big),$$ which means that $$A\cup(B\cap C)=(A\cup B)\cap (A\cup C).$$

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Reply to the OP's comments

I understand that the proof is a form of ∀x( P(x) ⟺ Q(x) ).

The meat of your proof (what you showed) has the structure P(k)⟺S(k)⟺T(k)⟺...⟺Q(k), where k is arbitrary; afterwards, you invoke universal generalisation to declare that ∀x( P(x) ⟺ Q(x) ).

There is no need to be showing ∀x( P(x) ⟺ S(x) ) or ∀x( S(x) ⟺ T(x) ), and we are certainly not showing that ∀x P(x) ⟺ ∀x S(x) ⟺ ... ⟺ ∀x Q(x).

Is the domain of discourse of $x$ any set containing $A,B,C?$

This implies that $x$ could stand for $B,$ which is of course incorrect; any superset of $A\cup B \cup C$ can be the discourse domain of your proof.

In '$∀x(P(x)→Q(x))$', since $x$ is universally quantified, '$x∈A$' is a proposition. What do you think?

No, in the proposition $\forall y\, V(y),$ the propositional function $V(y)$ is not a proposition. On the other hand, when $k$ is a constant, then $V(k)$ is a proposition.