Proof via induction for small-step semantics I'm doing a course in Computer Programming Languages and I'm trying to prove the following (roughly following Pierce's Types and Programming Languages book):

if $t \rightarrow^* t'$ then $if\; t\; then \; t2 \; else \; t3 \rightarrow^* if\; t'\; then \; t2 \; else \; t3$

I'm a little confused about where to start; as far as I know, I'm supposed to define a base case, then prove it via induction.

I've started my proof by assuming these base cases:

$P(\frac{}{true \rightarrow^* true})$ and 
$P(\frac{}{false \rightarrow^* false})$.

I'm stuck at this point, and I don't really how know to proceed. 

edit: I've added the syntax in an effort to clear things up.
The induction should be on the derivation of $t \to^* t'$:


If $t \to t'$ in one step, then we get the desired result by the E-IF rule.
Suppose $t \to^* t'$ because $t \to t'' \to^* t'$. By the E-IF rule we have
$$(\mathtt{if}\;t\;\mathtt{then}\;t_2\;\mathtt{else}\;t_3)
    \to
    (\mathtt{if}\;t''\;\mathtt{then}\;t_2\;\mathtt{else}\;t_3)$$
and by the induction hypothesis
$$(\mathtt{if}\;t''\;\mathtt{then}\;t_2\;\mathtt{else}\;t_3)
    \to^*
    (\mathtt{if}\;t'\;\mathtt{then}\;t_2\;\mathtt{else}\;t_3).$$
These two together give us the desired
$$(\mathtt{if}\;t\;\mathtt{then}\;t_2\;\mathtt{else}\;t_3)
    \to^*
    (\mathtt{if}\;t'\;\mathtt{then}\;t_2\;\mathtt{else}\;t_3).$$