Why does forcing not obey some simple rules of Propositional Calculus?

Question

In Cohen "Set Theory and the Continuum Hypothesis" Cohen states on page 118:

"Also forcing does not obey some simple rules of the propositional calculus. Thus p may force $\neg \neg A$ and yet not force A".

Question : Can anyone help me understand why this occurs, maybe by providing an example (and / or a reference where the reasoning is detailed)?

Answer 1

Just found this :

Is weak forcing a semantic relation?

and the solution is in the associated comments and Carl Mummert's answer.

So based on the above post an example would be :

Assume $\text{ not true } p \Vdash A$

The Cohen forcing definition for negation is : $$ p \Vdash \neg (\neg A) \iff \forall Q \supseteq p \text{ not true } Q \Vdash \neg A \tag{1}$$

So as Cohen forcing is intended to be negation complete (Cohen Lemma 3, p119) and consistent (Cohen Lemma 1, p118) there could be a q $\supset p$ : $q \Vdash A$ (q $\nsubseteq$ p as we are assuming $\text{ not true } p \Vdash A$).

If $ \exists q \supset p : q \Vdash A$ then $\forall Q \supseteq p \text{ not true } Q \Vdash \neg A$

So using (1) $p \Vdash \neg (\neg A)$ or equivalently $p \Vdash \neg \neg A$