Prove: Let α be a proposition containing only Boolean connectives ∧,∨. Then any assignment satisfying α must also satisfy f(α)

The question:

Let f be a mapping that takes as input a Boolean proposition (no quantifiers) and outputs the same proposition but with all ∧ symbols replaced by ∨. For example: $$f(x_1 ∧ (x_2 → ¬x_5) = (x_1 ∨ (x_2 → ¬x_5)$$ Prove: Let α be a proposition containing only Boolean connectives ∧,∨. Then any assignment satisfying α must also satisfy f(α).

I don't understand why this statement is true. I thought about a counter-example of α = β ∨ ¬β is a Tautology. Hence every assignment satisfies α, but f(α) = β ∧ ¬β is a Contradiction hence there is no assignment that satisfies f(α). what I'm missing here, and what is wrong with my thinking?

• Does $\beta \lor \neg \beta$ contain any conjunctions? Commented Jul 10 at 14:41
• Oh, I thought $f$ was mapping $\lor$ to $\land$. My bad. Commented Jul 10 at 14:42
• @PW_246 Not necessarily, I just thought about it as a counterexample, the full explanation might be done with Structural induction Commented Jul 10 at 14:43
• Your counterexample is not a counterexample, because $f(\alpha) = \beta \lor \neg \beta$, not $\beta \land \neg \beta$. Commented Jul 10 at 14:44
• @User33975329257439645 Yes, you can prove it by structural induction, but since there is the additional assumption that $\alpha$ contains only $\lor, \land$, you can also argue that since $(A \land B) \models (A \lor B)$, $f(\alpha)$ follows from $\alpha$, since both are strictly positive and $\land$ is stronger than $\lor$. Commented Jul 10 at 14:46