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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 $(2)$.

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    You are substituting $\neg P\vee Q$ for $P\to Q$ in $\neg R$ – 2017-01-23
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    @Casper I guess I get it: We substitute $R$ in $¬R \equiv ¬R$? It is very simple. I was thinking in an analogy of multiplying an equation by $-1$. – 2017-01-23
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    Yes; *substitute* zero or more occurrences of a sub-formula $\mathcal A$ with an equivalent formula $\mathcal B$ into a formula $\mathcal U$ to get a new formula $\mathcal U^*$ wich is equivalent to the original one." Here $\mathcal A$ is $P \to Q$ and $\mathcal B$ is $\lnot P \lor Q$ (the two are equivalent). – 2017-01-23

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$$¬(P\to Q)\equiv ¬(¬P\vee Q) \equiv ¬(¬P) \wedge ¬ Q \equiv P \wedge ¬ Q $$