1. A & b
2 A -> (D & G) /need to get D v Q

2. By simplification we get $A \wedge B\, \Rightarrow \,A$.
By modus ponems we get $D \wedge G$ from $A\, \Rightarrow \,\left( {D \wedge G} \right)$.
Then get D by simplification and then $D \vee G$ by disjunction.

3. Originally Posted by ambergarrett955
1. A & b
2 A -> (D & G) /need to get D v Q

3) A (conjunctive simplification)
4) D & G (using 3, assertion)
5) D (using 4, conjunctive simplification)
6) D v Q (using 5, addition)
QED.