Based on the comments I have added an omission to formula (1) which does affect the question I asked, which is hopefully now more specific.
In Enderton "A Mathematical Introduction to Logic" negation in a given structure M is defined as, with s the satisfaction function:
$$ \forall s \; ((M \models \neg A[s]) \; \Longleftrightarrow \; [\text{ not true } (M \models A[s]]) \tag{1}$$
If we take $$ (M \models A) := \; \forall s \; ((M \models A[s]) \; \tag{1a}$$ then:
$$(M \models \neg A) \; \Longrightarrow \; [\text{ not true } (M \models A)] \tag{1b}$$ but, $\mathbf{in \; general }$:
$$(M \models \neg A) \; \not\Longleftarrow \; [\text{ not true } (M \models A)] \tag{1c}$$
In deduction ($\vdash$), consistency is expressed as, for premises $p$, $p_1$, $p_2$:
$$ \text{not true } [(p_1 \vdash \neg A) \text{ AND } (p_2 \vdash A)]$$
or equivalently:
$$ (p_1 \vdash \neg A) \Longrightarrow [\text{ not true } (p_2 \vdash A)]$$
However consistency does NOT $\mathbf{in \; general }$ produce the right to left implication (taking $p_2$ as $p_1$):
$$ (p_1 \vdash \neg A) \not\Longleftarrow [\text{ not true } (p_1 \vdash A)] \tag{2}$$
which is equivalent to :
$$ [ \text{not true } (p_1 \vdash \neg A)] \; AND \; [\text{ not true } (p_1 \vdash A)] \tag{2a}$$
Indeed it is possible to find expressions that follow (2a), for example even though :
$$ \vdash (A \lor \neg A) \land (B \lor \neg B) $$
we can still have:
$$ [\text{not true } (\vdash A \land B)] \; AND \; [ \text{not true } (\vdash \neg (A \land B))] $$
or as another example:
$$ [\text{not true } (\vdash (A \lor B \implies A \land B))] \; AND \; [ \text{not true } (\vdash \neg (A \lor B \implies A \land B))] $$
So my question is: Is it possible to add some restrictions to make (1c) and (2) true. I thought it might be possible since in Cohen's book "Set Theory and the Continuum Hypothesis" he appears to be making definitions for a new operator $\Vdash$ that seem to just assume the following property $$ (p \Vdash \neg A) \Longleftarrow [ \text{ not true } (p \Vdash A)] \tag{3}$$ I think Cohen starts by assuming all formulae are false, then is careful about the ones chosen to be true, by like (3) assuming if a formula can't be shown to be true then it stays false, but I don't know why that is OK. So is there a general way of taking all the formulae that can't be shown to be true or not true in (1c) and (2) and carefully assign which are to be true and which are to be false, so every formula is shown to be true or not true, i.e. like (3), without causing any issues ?