Statement: If you study, then you can get a diploma. If you don't study then you can still get a diploma.
$S \to D$
0 1 0
0 1 1
1 0 0
1 1 1
$\text{not}(S)\to D$
1 0 0
1 1 1
0 1 0
0 1 1
When I use truth table I get conflict(not all 1's), but I'm not sure that this means I can deduce anything from that statement.
Please correct me if I'm wrong.
You can deduce (in classical propositional logic) that 'you can get a diploma' is true.
Proof. Keeping in line with your notation we have that $S \implies D$ and $\neg S \implies D$. Now the Constructive Dilemma is the following rule $$ (( p \implies q ) \wedge (r \implies s) \wedge (p \vee r)) \vdash q \vee s $$ Setting $p := S, r := \neg S$ and $q := s := D$ this yields $$ (( S \implies D ) \wedge ( \neg S \implies D) \wedge (S \vee \neg S)) \vdash D \vee D. $$
Now, by Tertium non datur (Law of Excluded Middle), $\vdash S \vee \neg S$ and hence $$ (( S \implies D ) \wedge ( \neg S \implies D)) \vdash D \vee D. $$ Since $\vdash (D \vee D) \iff D$, this may be further simplified to $$ (( S \implies D ) \wedge ( \neg S \implies D)) \vdash D. $$