$P(B|\{X=i\}) = a$ equivalent to $X=i⟹P(B)=a$? 
Question: Let $X$ be a random variable then is the statement $\mathbb{P}(B|\{X=i\})$ = 0.6(random number), equivalent of saying: $X = i \implies \mathbb{P}(B) = 0.6$?


Also say in the case A and B are conditionally independent if we condition on the event $\{ X = i \}$, which is $\mathbb{P}(A \cap B|\{ X = i \}) =\mathbb{P}(A|\{ X = i \}\mathbb{P}(B|\{ X = i\})$. Is it equivalent to:
$X = i \implies \mathbb{P}(A\cap B) =\mathbb{P}(A)\mathbb{P}(B)?$

I don't know what conditional probability means from a measure theory perspective, but I was taught in undergraduate probably that $\mathbb{P}(A|B)$ where A and B are events is defined as $\frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}$. However, treating it as such a definition is somewhat weird when I write proofs with regard to Markov chains, so can I just treat it as the first order logic implication as above?
If I can, I also wonder if it is possible to write it in this first order logic form for some events B, $\mathbb{P}(A|B)$, where B does not come from a preimage of what a random variable.
 A: If you would have said something like:
"Based on the info that random variable $X$ has taken value $i$ we conclude that the probability of occurrence of event $B$ is now $0.6$"
then I would agree, so IMV the intuition on this is okay.
But I disagree with using a logical implication for that. Firstly its premisse is false (because $X$ is a function and $i$ is not) so that the implication is always true no matter what is implied.
If you repair this by taking something like $X(\omega)=i$ then still things are not okay because $\mathbb P(B)$ is a fixed real number not depending on $\omega$ or $i$.
Let's just do it without that implication. Our intuition does not really need it.
A: $$P(B|A)=k\tag1$$ $$A \text{ happens}\implies P(B)=k\tag2$$
It is false that for every pair of events $A$ and $B$ and every $k\in[0,1],$ statements $(1)$ and $(2)$ are equivalent.
A counterexample is a toss of a fair coin, with $A$ and $B$ being the event of a head and of a tail, respectively, and with $k$ being $0.5;$ here, statement $(1)$ is false while statement $(2)$ is true.

Addendum

OP: Could you elaborate on the counterexample? I don't really understand what you mean.

When you plug the counterexample into equation $(1),$ LHS $=0\ne$ RHS, so statement $(1)$ is False; when you plug the counterexample into implication $(2),$ its consequent(conclusion) is True, so, regardless of its antecedent(premise)'s truth value, statement $(2)$ is True.
