Are the following two statements logically equivalent? 
*

*If there exists A such that B, then C.

*There exists A such that if B, then C.

I am having a hard time telling whether these two statements are logically equivalent.
(I have an easier time understanding 1 than 2, so if they are equivalent, it will make me feel more comfortable when I read statements like 2.) Thank you!
Edit: I have this question because I do not know how to interpret a theorem (Theorem 1.1) by Candes and Tao (2010): https://arxiv.org/pdf/0903.1476.pdf. I copy the entire theorem here:

Theorem $1.1$ (Matrix completion I) Let $M \in \mathbb{R}^{n_{1}
> \times n_{2}}$ be a fixed matrix of rank $r=O(1)$ obeying the strong
incoherence property with parameter $\mu$. Write $n:=\max \left(n_{1},
> n_{2}\right)$. Suppose we observe $m$ entries of $M$ with locations
sampled uniformly at random. Then there is a positive numerical
constant $C$ such that if $$ m \geq C \mu^{4} n(\log n)^{2}, $$ then
$M$ is the unique solution to (1.3) with probability at least
$1-n^{-3}$. In other words: with high probability, nuclear-norm
minimization recovers all the entries of $M$ with no error.

If I follow the explanation by ryang (thank you for answering), then "M is unique solution..." if $m \geq C \mu^{4} n(\log n)^{2}$ for all numerical constant $C$. But this does not make sense since $m \geq C \mu^{4} n(\log n)^{2}$ clearly does not hold for all $C$. What am I missing here?
 A: If there was a day where the ground was wet, it was raining.
There was a day where if the ground was wet, it was raining.
Suppose it was dry the first day, then the next day someone sprayed the ground with a hose. Then the first statement is false (the ground was wet on the second day but it never rained) but the second is true (on the first day the ground wasn't wet, so the conditional is satisfied).
In logical terms, $[\exists A B(A)]\rightarrow C$ is actually equivalent to $\forall A[B(A)\rightarrow C]$. The intuition here is that the existence of any $A$ with $B(A)$ implies $C$. Thus, every individual $A$ has the property that $B(A)$ implies $C$.
A: In some sense, "there exists" statements are harder to make true than implications. Just set the domain of discourse to be empty, and you'll immediately see that they are not equivalent: the first one is true because the antecedent is always false, while the second one is false because nothing exists.
A: 

*

*If there exists A such that B, then C.

*There exists A such that if B, then C.

(B depends on A, but C does not.)

To be clear, A is a variable, B is a propositional function of the variable A, and C is a proposition. This notation is confusing, so let's write $$a,\;B(a),\;C$$ instead. Then, formally:

*

*$$\color{green}{∃a\,B(a)→C}\quad\equiv\quad ∀a\;\big(B(a)→C\big)$$
$\color{green}{\text{$C$ is true if some $a$ satisfies $B(a),$ that is, if any of $B(a_1),B(a_2),B(a_3),$ etc. is true.}}$
Every $a$ that satisfies $B(a)$ makes $C$ true.


*$$∃a\;\big(B(a)→C\big)\quad\equiv\quad \color{green}{∀a\,B(a)→C}$$
There is some $a$ such that if it satisfies $B(a)$ then $C$ is true.
$\color{green}{\text{$C$ is true if every $a$ satisfies $B(a),$ that is, if $B(a_1),B(a_2),B(a_3),$ etc. are all true.}}$
Several parses may be needed to convince oneself of these two equivalences.
Comparing the green sentences: $(1)$ logically implies $(2),$ but not vice versa. Here's a counterexample: \begin{align}\text{domain }&:=&\mathbb R\\B(a)&:=&a>0\\C&:=&3=7.\end{align}
