Logical Argumentation of $[(\phi \lor \psi) \implies \theta] \iff [(\phi \implies \theta) \wedge (\psi \implies \theta)]$

Question

I'm trying to practice arguing logical equivalence. I know how to do this via truth tables, or by some applications of contrapositives, but I'd like to get a handle on logical argumentation.

My biggest question is: is my argument is valid? Aside from that, all feedback/critique is welcomed (did I make my assumptions clear? is this too verbose? etc)

Currently I have written the following argument:

---

Claim

$[(\phi \lor \psi) \implies \theta] \iff [(\phi \implies \theta) \wedge (\psi \implies \theta)]$

Argument/Proof

$\Rightarrow$

Given $\phi$, then $\phi \lor \psi$ holds. Since $\phi \lor \psi$ holds when $\phi$, we can conclude $\theta$. Thus, $\phi \implies \theta$

Given $\psi$, then $\phi \lor \psi$ holds. Since $\phi \lor \psi$ holds when $\psi$, we can conclude $\theta$. Thus, $\psi \implies \theta$

Hence, $\phi\implies\theta$ and $\psi\implies\theta$

Thefore, $[(\phi \lor \psi) \implies \theta] \implies [(\phi \implies \theta) \wedge (\psi \implies \theta)]$

$\Leftarrow$

Given $\phi$, $(\phi\implies\theta)\wedge(\psi\implies\theta)$, we can conclude $\theta$ from $\phi\implies\theta$ when $\phi$

Given $\psi$, $(\phi\implies\theta)\wedge(\psi\implies\theta)$, we can conclude $\theta$ from $\psi\implies\theta$ when $\psi$

If we can deduce $\theta$ from $\phi$, or deduce $\theta$ from $\psi$, we can deduce $\theta$ from $\phi$ or $\psi$. That is, $(\phi\lor\psi)\implies\theta$

Therefore, $[(\phi \implies \theta) \wedge (\psi \implies \theta)] \implies [(\phi \lor \psi) \implies \theta]$

Answer 1

Your argument is (syntactically) valid and perfectly clear. For beginning logic classes, professors (sometimes) ask that people list out the particular rules of inference used at each line (Fitch Style). But if your professor does not ask, I would not include.