Can we logically analyze mathematical theorems as if-then statements? Many theorems in math have an if-then form. For example: "If a polynomial is of $n^{th}$ degree, then it has $n$ roots. In my other question, I learned that in order to analyze statements using truth tables, the statements must be completely independent. However, I'm not sure that anything in math is independent. Everything can be proven from the axioms as far as I know, so if $p \implies q$, the truth value of $p$ automatically determines the truth value of $q$. So if the statement "$f$ is polynomial is a degree $5$" automatically makes the statement $f$ has $5$ roots", the truth table seems nonsensical. Do mathematical if-then statements have anything to do with the classical if-then statements from logic? 
$$p \implies q$$
$$\begin{array}{|c|c|c|}
\hline
p&q&p\implies q\\ \hline
T&T&T\\
T&F&F\\
F&T&T\\
F&F&T\\\hline
\end{array}$$
 A: 
Do mathematical if-then statements have anything to do with the classical if-then statements from logic?

Yes; they use "the same" if-then.

Everything can be proven from the axioms 

or, more precisely, in a mathematical theory every theorem is proved from the axioms.
This amount to:

if the axioms are true, then also the theorems are.

What happens if the axioms are not true? Well, the conditional still holds, but we are losing our time with a "wrong" mathematical theory.

A well-know quote from Bertrand Russell, The Principles of Mathematics page 3:

  
*
  
*Pure Mathematics is the class of all propositions of the form “$p$ implies $q$”, where $p$ and $q$ are propositions containing one or more variables,
  the same in the two propositions, and neither $p$ nor $q$ contains any constants
  except logical constants.
  


Examples of arithmetical theorems that are conditionals with false antecedent.
1) "if $2$ is odd, then $2=1$".
Proof
We define : $Even(n) := \exists z (n = z \times SS0)$.
Finally, we define: $Odd(x) := \lnot Even(x)$.
From Peano axioms it is easily proved that: $Even(SS0)$, i.e. $Even(2)$.
Thus, having proved $\lnot Odd(2)$, using the tautology:

$\lnot A \to (A \to B)$,

with $Odd(2)$ in place of $A$ and $2=1$ in place of $B$, by modus ponens we conclude with:


$Odd(2) \to 2=1$.


2) "if $2$ is odd, then $2=2$".
Proof
From equality axiom: $\forall x (x=x)$, we get: $2=2$.
Thus, using the tautology: 

$A \to (B \to A)$,

with $2=2$ in place of $A$ and $Odd(2)$ in place of $B$, by modus ponens we conclude with:


$Odd(2) \to 2=2$.


