Prove "If $\{R, Q\} \vDash P$, then $¬P \vdash \lnot Q \lor \lnot R$". Using the book Dirk van Dalen. "Logic and Structure (Universitext)" as reference text.
Decide whether the following statement is true or false, justifying the answer.

If $\{R, Q\} \vDash P$, then $¬P \vdash \lnot Q \lor \lnot R$.

I claim this statement is true.
Proof:
Assume $[[R]]_v=[[Q]]_v=1 \Rightarrow [[P]]_v=1$.
Let v an arbitrary valuation.
By completeness theorem,

$¬P \vDash \lnot Q \lor \lnot R \Rightarrow ¬P \vdash \lnot Q \lor \lnot R$

So, we can now try to prove $¬P \vDash \lnot Q \lor \lnot R$.
Assume $[[\lnot P]]_v=1$ (*).

*

*Assume $\lnot ([[\lnot Q \lor \lnot R]]_v = 1)$

*

*Assume $[[R]]_v=[[Q]]_v=1$

*$[[P]]_v=1$

*$[[P]]_v=0$ (From (*))

*$\bot$



*$\lnot([[R]]_v=[[Q]]_v=1)$

*$[[\lnot Q]]_v=1 \lor [[\lnot R]]_v=1$

*$[[\lnot Q \lor \lnot R]]_v = 1$

*$\bot$
$[[\lnot Q \lor \lnot R]]_v = 1$
$[[¬P]]_v=1 \Rightarrow [[\lnot Q \lor \lnot R]]_v=1$
$¬P \vDash \lnot Q \lor \lnot R$
I will appreciate if someone could give some insight into the general structure of the proof and I would like to know whether my attempt is correct.

The basic proof plan is correct ... but it is a little werd to use propositional logic notation for a meta-proof.
For example, you say things like

$\lnot([[R]]_v=[[Q]]_v=1)$

and

$[[\lnot Q ]]_v =1 \lor [[\lnot R]]_v = 1$

so now you are using propositional logic operators to make your meta-logical claims, which is very confusing.
So, I would suggest to just stick with English, just as you would with any mathematical proof, and instead say:

it is not the case that both $[[R]]_v =1$ and $[[Q]]_v =1$

and

either $[[\neg Q]]_v =1$ or $[[\neg R]]_v =1$

