I'm really stuck on a logical equivalance question using substitution and transitive property. The question i'm trying to prove is;
$a \implies b \equiv \neg (a \land \neg b)$
and so far i have
$a \implies b \equiv$
\begin{align*} &\neg b \implies \neg a \\ &\neg \neg a \implies \neg \neg b \\ &b \lor ¬a \\ &\neg a \land b \\ &\neg(a \land \neg b) \end{align*}
I'm just second guessing myself because I feel like all i'm doing is flipping the inverse of $a$ and $b$ and just declaring equivalence between them without having correct substitution or transitive property. I understand simpler examples where proving C if A implies B and B implies C etc, but this particular example is really confusing me, just wondering if anyone could help clear this up. Thanks in advance.
First of all, in your 'derivation' the fourth line $\neg a \lor b$ is not equivalent to either the third line r the fifth line. So that's certainly not correct.
Also, while the first two steps are correct applications of Contraposition, going from $\neg \neg a \to \neg \neg b$ to $b \lor \neg a$ is not something I recognize as the application some elementary equivalence principle, because typically the equivalence dealing with the material implication is:
Implication
$P \rightarrow Q \Leftrightarrow \neg P \lor Q$
So, you get:
$$A \to B \overset{Implication}{\Leftrightarrow}$$
$$\neg A \lor B \overset{Double \ Negatoin}{\Leftrightarrow}$$
$$\neg A \lor \neg \neg B \overset{DeMorgan}{\Leftrightarrow}$$
$$\neg (A \land \neg B)$$