Are truth tables a valid method to prove an iff statement? I recently had a homework assignment returned to me (for a Differential Geometry course, undergrad level) in which my instructor wrote "You cannot use truth tables to prove an if and only if statement". I used truth tables to prove two questions from O'Neill's "Elementary Differential Geometry (Revised 2nd Edition)": problem #2(a&b)  of Section 5.5 and problem #1 of section 5.6. 
My question is as simple as the title. I only want to know if truth tables can or cannot be used to prove an if and only if statement. Most sources I've checked say you can, but at the same time, my instructor says you can't.
EDIT: Question: Prove a curve $\alpha$ in M is a straight line of R3 if and only if $\alpha$ is both geodesic and asymptotic.
My answer was: 
Statement 1: By definition, a curve $\alpha$ $\in$ M $\subset$R3 is asymptotic if $\alpha''$ is tangent to M, where S($\alpha$)=-U' and U$\cdot$ $\alpha''$=0 where U'$\cdot$$\alpha'$=0 if and only if U$\cdot$$\alpha''$=0.
Statement 2: By definition, a curve $\alpha$ $\in$ M $\subset$R3is geodesic if $\alpha''$ is always normal to M, and 2$\alpha'$$\cdot$$\alpha''$=0 implies $\alpha''$=0.
Statement 3: By definition, a curve $\alpha$ $\in$ M $\subset$R3 is a straight line iff and only iff $\alpha$(t)=p+tq, such that $\alpha'$=q and $\alpha''$=0. 
my truth table showed (statement 1 AND statement 2), then staement 3 if and only if (statement 1 AND staement 2).
 A: Your set-up is valid, but the math falters a bit. Your final conclusion in (2) is incorrect. So the correct logic is this: $\alpha''=0$ if and only if its component normal to $M$ and its component tangent to $M$ are both $0$. 

(1) says $\alpha''$ is orthogonal to $U$ and (2) says $\alpha''$ is parallel to $U$. Thus, (1) and (2) is equivalent to $\alpha''=0$.

You don't need truth tables. Just logic and math in words. Your teacher may have complained that the key idea was unclear.
A: Suppose you want to show that $\phi$ is true iff $\psi$ is true. You can make a truth-table for the propositional letters occurring in either $\phi$ or $\psi$, and check whether $(\phi \equiv \psi)$ gets assigned $\top$ by all rows of the truth-table. If the equivalence between $\phi$ and $\psi$ is a truth of propositional logic, this decision procedure is guaranteed to succeed. 
If, however, $(\phi \equiv \psi)$ is not a truth of propositional logic, but only follows from a set of axioms $\Gamma$, then you can show $\Gamma \rightarrow (\phi \equiv \psi)$ using truth-tables instead. This wouldn't be the best way of showing the equivalence of $\phi$ and $\psi$ given $\Gamma$, but it would work. From your comments I gather that this is exactly what you've done.
A: There exist mathematical propositions which are currently undecided.  If a proof comes as required for a statement, one might claim that the default assumption lies in that such a proposition is currently undecided by the intended reader of the statement.  If so, then two-valued truth tables do NOT suffice, since you'd have to write out the truth tables in the case when the proposition is undecided.  This entails using at least a three-valued logic, but then you'll have to decided which operation you want to use for logical equivalence.  So, such a truth table won't show that the proposition holds true, because the definition of logical equivalence will get contested.
There also exists the problem that if you were to formalize everything here you'd end up needing quantifiers, and consequently truth tables might come as impossible to construct.
