Formal proof structure for $\forall n \in \mathbb{N}, P(n) \rightarrow \forall n \in \mathbb{N}, Q(n)$

Question

I'm used to proving universal quantification claims (i.e. $\forall n \in \mathbb{N}, [P(n) \rightarrow Q(n)]$) by: Assuming an arbitrary number in the naturals, assuming the antecdent $P(n)$, doing the mechanical portion of the proof to get $Q(n)$, conclude. However, how do I go about proving statements of the form:

$$\forall n \in \mathbb{N}, P(n) \rightarrow \forall n \in \mathbb{N}, Q(n)$$

I would be very greatful if someone could provide an explanation on how to prove such a statement. Thank you!

Answer 1

Just two of several possibilities...

Direct proof:

1. Suppose $\forall n \in \mathbb{N}: P(n)$

2. Suppose $x\in \mathbb{N}$

3. Prove $Q(x)$

4. Conclusion: $\forall n \in \mathbb{N}: Q(n)$

5. Conclusion: $\forall n \in \mathbb{N}: P(n) \rightarrow \forall n \in \mathbb{N}: Q(n)$

Indirect proof by contradiction:

1. Suppose $\forall n \in \mathbb{N}: P(n)$

2. Suppose that $x\in \mathbb{N}\land\neg Q(x)$

3. Obtain a contradiction of the form $A\land\neg A$, or $A\iff\neg A$

4. Conclusion: $\neg \exists n\in\mathbb{N}: \neg Q(n)$, or equivalently, $\forall n\in\mathbb{N}: Q(n)$

5. Conclusion: $\forall n \in \mathbb{N}: P(n) \rightarrow \forall n \in \mathbb{N}: Q(n)$