Rules of distribution of quantifiers over conditional and biconditional

Which of the following propositional logic statements are true and why?

1. $(∀x(P(x)⟹Q(x)))⟹((∀xP(x))⟹(∀xQ(x)))$
2. $(∀x(P(x))⟹∀x(Q(x)))⟹(∀x(P(x)⟹Q(x)))$
3. $(∀x(P(x))⇔(∀x(Q(x))))⟹(∀x(P(x)⇔Q(x)))$
4. $(∀x(P(x)⇔Q(x)))⟹(∀x(P(x))⇔(∀x(Q(x))))$

5. Are their any standard laws/rules of distribution of universal quantifier over conditional and binconditional that can help me solve this?

6. Also rules for distribution of existential quantifier over conditional and binconditional?

Recently I came across distribution of quantifiers over $\vee$ and $\wedge$, which gave set theoretic interpretation of them as follows:

• $((∀x)G(x)∨ (∀x)H(x))→ (∀x)(F(x)∨ G(x))$

In set theoretic terms, if we have that $(f(G) = D ∨ f(H) = D)$, then we have $(f(G) ∪ f(H)) = D$

• $(∃x)(G(x)∧ H(x))→((∃x)G(x)∧ (∃x)H(x))$

In set theoretic terms, if we have that $(f(G) ∩ f(H)) ≥ 1$, then we have $(f(G) ≥ 1 ∧ f(H) ≥ 1)$

Can we say similar for distribution of quantifiers over conditional and biconditional (just to bring in more clarity)?

• The above formulas are not propositional, since they use quantifiers. They are just formulas of first-order logic. Commented Jan 1, 2016 at 21:23
• Do you want me to change the title to something else? Commented Jan 1, 2016 at 21:25
• Yes, the title is misleading, as well as the tag "propositional calculus". Commented Jan 1, 2016 at 21:27
• Does the new one makes sense "Rules of distribution of quantifiers over conditional and biconditional makes sense?" Commented Jan 1, 2016 at 21:31
• Now the title is ok for me. Commented Jan 1, 2016 at 21:33

Formulas (1) and (4) are valid, i.e. they are true in every first-order $\mathcal{L}$-structure.

Formulas (2) and (3) are not valid, i.e. there exists a $\mathcal{L}$-structure in which they are not true. For instance, take the $\mathcal{L}$-structure $\mathcal{N}$ whose domain is $\mathbb{N}$ and whose interpretation of $P$ is $2\mathbb{N}$ (the set of even natural numbers), and whose interpretation of $Q$ is $\mathbb{N} \smallsetminus 2\mathbb{N}$ (the set of odd natural numbers). You have that the formula $\forall xP(x) \Rightarrow \forall x Q(x)$ is vacuously true in $\mathcal{N}$ (it claims that "if every natural number is even then every natural number is odd"), but the formula $\forall x(P(x) \Rightarrow Q(x))$ is false in $\mathcal{N}$ (it claims that "for every natural number, if it is even then it is odd"), therefore your formula (2) is false in $\mathcal{N}$. Similarly for the formula (3), since $A \Leftrightarrow B$ is equivalent to $(A \Rightarrow B) \land (B \Rightarrow A)$.

In general, when one talks about distributivity of something over something else (for instance, distributivity of $\land$ over $\lor$), one means that two formulas are logically equivalent. With this meaning, the answer to your question "Does the universal quantifier distribute over conditional or biconditional?" is negative since the formula $\forall xP(x) \Rightarrow \forall x Q(x)$ is not logically equivalent to the formula $\forall x(P(x) \Rightarrow Q(x))$ (your formula (1) is valid, but your formula (2) is not valid), and similarly the formula $\forall xP(x) \Leftrightarrow \forall x Q(x)$ is not logically equivalent to formula $\forall x(P(x) \Leftrightarrow Q(x))$ (your formula (4) is valid, but your formula (3) is not valid).

• Well I have not taken a dedicated course on mathematical strucutres or read any books specifically on mathematical structures. So I am finding it difficult to deal with $L$-structure thingy... Commented Jan 2, 2016 at 9:10
• ...However,if interprete $P(x)$ as "x is a boy" & "$Q(x)$ is clever". Then, $(∀x(P(x)⇔Q(x)))$ means "If any $x$ is a boy then he is clever & vice versa". This definitely implies $(∀x(P(x))⇔(∀x(Q(x))))$, "If all $x$ are boys then all are clever & vice versa". But converse does not seem valid as $(∀x(P(x))⇔(∀x(Q(x))))$ puts restriction: "if all $x$ are boys". $(∀x(P(x)⇔Q(x)))$ requires that "for any $x$, if $x$ is a boy", but does not require that "all $x$ are boys", so there may be a person who is not boy, but still $(∀x(P(x)⇔Q(x)))$ will be valid but $(∀x(P(x))⇔(∀x(Q(x))))$ will not be... Commented Jan 2, 2016 at 9:11
• ...So 3 seems invalid to me, while 4 seems valid. Also what about distributing existential quantifier? Commented Jan 2, 2016 at 9:11
• @PardonMeForMySuperPoorMaths: Sorry, there was a typo in my answer, now I fixed it. Your formula (4) is the "biconditional version" of your formula (1), and your formula (3) is the "biconditional version" of your formula (2). Commented Jan 2, 2016 at 10:14
• any comment about the same but involving existential quantifier? Commented Jan 2, 2016 at 12:50