I want to deduce that
$$¬(P\to Q)\equiv P\wedge¬Q$$
In the book I'm reading, he suggests to use:
$$(P\to Q)\equiv ¬P\vee Q\tag{1}$$
and substitute to get:
$$¬(P\to Q)\equiv ¬(¬P\vee Q)\tag{2}$$
I am able to deduce it by negating both sides of $(1)$, which (with De Morgan's laws) yields precisely the desired conclusion. My problem is that I am negating both sides but this is not really an admissible operation, I should use the substitution principle, which is:
Substitution Principle: If $\mathfrak U$ is a formula contaning a component formula $\mathfrak a$ and if $\mathfrak U^*$ is a formula obtained from $\mathfrak U$ by substituting a formula $\mathfrak b$ for one of more occurrences of $\mathfrak a$ in $\mathfrak U$, and if $\mathfrak a \equiv \mathfrak b$ is a tautology, then $\mathfrak U \equiv \mathfrak U ^*$ is a tautology.
I really don't see what substitution is being made to transform $(1)$ in $(2)$.
$$¬(P\to Q)\equiv ¬(¬P\vee Q) \equiv ¬(¬P) \wedge ¬ Q \equiv P \wedge ¬ Q $$