To express quantification of propositional function using $\land , \lor , \lnot$

Question

$\forall x((x \neq 3) \rightarrow P(x))\equiv \forall x((x=3) \lor P(x))) $ ,where domain of P(x) consists of intergers 1,2,3,4,and 5.

The given answer is:$(P(1)\land P(2)\land P(4)\land P(5))$,

Since $\forall x((x \neq 3) \rightarrow P(x))\equiv \forall x((x=3) \lor P(x)) $,

I'm confused why not the answer be:

$((x=3 \lor P(1)) \land ... \land (x=3 \lor P(5))$

Answer 1

When we instantiate $\forall x ((x=3) \lor P(x))$ we get:

$((1 =3) \lor P(1)) \land ((2=3) \lor P(2)) \land ((3=3) \lor P(3)) \land ((4=3) \lor P(4)) \land ((5=3) \lor P(5))$.

But $(1=3) \equiv \text F$ and $(F \lor P(1)) \equiv P(1)$. The same for $2,4,5$.

We have also $(3=3) \equiv \text T$ and $(T \lor P(3)) \equiv T$.

In conclusion, we get:

$P(1) \land P(2) \land T \land P(4) \land P(5)$

i.e. $P(1) \land P(2) \land P(4) \land P(5)$, as expected.