Let $P$ be the statement: $\forall x \in \mathbb{R}, \exists y \in \mathbb{R}$ such that $y \geq x \,\, \wedge \,\, |f(y)| \geq 1$
Let $Q$ be the statement: $\exists x \in \mathbb{R}, \forall y \in \mathbb{R}$ such that $y \geq x \Rightarrow \, |f(y)| \geq 1$.
Prove, or disprove by counterexample, that (a) $P \Rightarrow Q$ and (b) $Q \Rightarrow P$.
So I've noted that the inverse of the statement $Q$ can be represented as: $\forall x \in \mathbb{R}, \exists y \in \mathbb{R}$ such that $y \geq x \,\, \wedge \,\, |f(y)| < 1$, which gets me to a form closer to $P$, but I'm not sure how to complete the proofs required from there. Is that even the right direction to try?