Can I prove that $A\Rightarrow (B\wedge C)$ by proving that $\neg B\Rightarrow\neg A$?

Solution 1:

1. Re: your first question: unfortunately, your suggestion is invalid. With the counterexample $(A,B,C)=(\text{True, True, False}),$ $$\neg B\Rightarrow\neg A$$ is true but $$\neg(B\wedge C)\Rightarrow\neg A$$ false.

2. Your last section is a proof of $$(¬(B ∧ C) \Rightarrow¬A) \Rightarrow(¬B \Rightarrow ¬A),$$ rather than a proof of $$(\neg B\Rightarrow\neg A)\Rightarrow(A\Rightarrow(B\wedge C)).$$

   While the latter is not logically true (my above counterexample shows that it can be false), the former is a tautology, i.e., is logically true.

Addendum

EDIT: I understand from the answers that my reasoning is false. Suppose nonetheless that I still want I to show $$\neg(B\wedge C)\Rightarrow \neg A$$ How do I do it?

Remember, you are proving a statement of this form, not the proposition as it currently stands. For example, does $B$ stand for $\exists x{\in}\mathbb Z\;x^2=8,$ or does it symbolise a compound proposition? That informs your proof strategy.

Solution 2:

If $A$ and $B$ are true while $C$ is false, then $\lnot B\to\lnot A$ is true and $\lnot (B\land C)$ is true, meaning $\lnot (B\land C)\to \lnot A$ is false; a true statement cannot imply a false statement.