Predicate Calculus with Sets - Question about use of an axiom Greets again StackExchange,
I am watching an online lecture, and I believe that my instructor has misused an axiom. Is my concern warranted?
$$\begin{align*}
\text{Given:}&  {P \subseteq (Q \cap R)}\\
&{(Q \cup S) \subseteq T}\\
&{x \in (P \cup S)  }\\
\text{Prove:}& {x, \in T}
\end{align*}$$
I will reference a membership function, P(x), so onward then to the proof.
Translating my givens into Predicate Calculus
$$\begin{align*}
\text{Given:}&\\
&1.\ {\forall x P(x) \rightarrow [Q(x)\wedge R(x)]}\\
&2.\ {\forall x [Q(x) \vee S(x)] \rightarrow T(x)}\\
&3.\ {\exists x [P(x) \vee S(x)]  }
\end{align*}$$
So now comes my problem. She claims in the following line, that she just used straight simplification on the R.H.S. of  line 1 to get Q(x) by itself as shown:
$$\qquad\quad 4.\  {\forall x [Q(x) \wedge R(x)] \rightarrow Q(x)} $$
Wouldn't she need To have $P(x)$ first, and then use modus ponens to get the right side of that implication first, then use simplification? Is my understanding of simplification incorrect? Can you just simplify any conjoined statement at will?
Thank you. If you could also venture an answer at the answer to the proof I would appreciate it, because without this step she goes over, I am at a loss on how to solve.

If $x \in P$ then $x \in Q$. But then, if $x \in P \cup S$ we have $x \in Q \cup S \subseteq T$.
