Difference due to scope of quantifiers

Question

I'm trying to understand the difference between the scopes of quantifiers in the following statements and how that impacts taking negations and contrapositives.

Statement 1: $\quad \exists M \in \mathbb{R}^+ \quad \forall x\in \mathbb{R}^+ \quad \big(x\geq M \implies f(x)>0 \big).$

I believe I know how to take the negation and contrapositive of statement 1:

• Negation: $\quad \forall M \in \mathbb{R}^+ \quad \exists x\in \mathbb{R}^+ \quad \big(x\geq M \land f(x)\leq0 \big).$
• Contrapositive: $\quad \exists M \in \mathbb{R}^+ \quad \forall x\in \mathbb{R}^+ \quad \big(f(x)\leq0 \implies x<M \big).$

Statement 2: $\quad \big(\exists M \in \mathbb{R}^+ \quad \forall x\in \mathbb{R}^+ \quad x\geq M\big) \implies f(x)>0 .$

Given statement 2 as above, my questions are:

1. Is there a difference between statement 2 and statement 1, and if so, what is it? An intuitive approach would be appreciated if possible.
2. If there is a difference, how does that impact the negation and contrapositive of statement 2?
3. Finally, if there are any mistakes in my negation and contrapositive of statememt 1, I would appreciate corrections.

Answer 1

Statement 1 doesn't have a contrapositive.

Only conditionals have contrapositives, and 1 isn't a conditional.

More carefully, in a formal context, only formulae whose main logical operator is the conditional (i.e. which are instances of the schema $A \to B$) have contrapositives (the corresponding instance of $\neg B \to \neg A$). And the main operator of 1 is the initial quantifier, so the notion of contraposition doesn't directly apply.

However, if you have a quantified formula $Q_1Q_2\ldots Q_n(A \to B)$ then that will be equivalent to what you get by contraposing $(A \to B)$, i.e. to $Q_1Q_2\ldots Q_n(\neg B \to \neg A)$ but of course leaving the quantifier prefix untouched!

Statement 2 is simply ill-formed in any sensible syntax.

Answer 2

Statement 1: $\quad \exists M \in \mathbb{R}^+ \quad \forall x\in \mathbb{R}^+ \quad \big(x\geq M \implies f(x)>0 \big).$

Statement 2: $\quad \big(\exists M \in \mathbb{R}^+ \quad \forall x\in \mathbb{R}^+ \quad x\geq M\big) \implies f(x)>0 .$

1. Is there a difference between statement 2 and statement 1, and if so, what is it? An intuitive approach would be appreciated if possible.

Good observation that Formulae 1 & 2 have different negations and different contrapositives, and thus different meanings!

Side note: Formula 2 is more clearly written as $$\quad \big(\exists M \in \mathbb{R}^+ \quad \forall y\in \mathbb{R}^+ \quad y\geq M\big) \implies f(x)>0$$ (notice that its variable $x$ is actually free and thus requires context), so is called an open formula instead of a statement.

Now, consider the logic formula \begin{gather}\Big(∀x\:Px\Big)\to Qy \quad↔\quad ∃x\:\Big(Px\to Qy\Big).\tag1\end{gather} If $(1)$'s RHS is false, then $∀x\:\Big(Px\land \lnot Qy\Big),$ so $Px$ is universally true and $\lnot Qy$ true, so $Px$ is universally true while $Qy$ false, so $(1)$'s LHS is false; by contrapositive and since we have been abstractly inferring, $(1)$'s LHS logically implies $(1)$'s RHS. On the other hand, if $(1)$'s LHS is false, then $Px$ is universally true and $Qy$ false, so $(Px\land \lnot Qy)$ is universally true, so $(Px\to Qy)$ is universally false, so $(1)$'s RHS is false; by contrapositive and since we have been abstractly inferring, $(1)$'s RHS logically implies $(1)$'s LHS. Hence, $(1)$'s LHS and RHS are logically equivalent, that is, \begin{gather}\Big(∀x\:Px\Big)\to Qy \quad\equiv\quad ∃x\:\Big(Px\to Qy\Big).\end{gather} Similarly, \begin{gather}\Big(∃x\:Px\Big)\to Qy \quad\equiv\quad ∀x\:\Big(Px\to Qy\Big).\end{gather}

2. If there is a difference, how does that impact the negation and contrapositive of statement 2?
3. Finally, if there are any mistakes in my negation and contrapositive of statement 1, I would appreciate corrections.

Your negation and contrapositive of Formula $(1)$ are both correct; just do the same for Formula $(2)$ (the procedure is more straightforward as the main connective ‘${\Rightarrow}$’ is not being quantified).