• December 5th 2007, 07:49 PM
ambergarrett955
1. A & b
2 A -> (D & G) /need to get D v Q
• December 6th 2007, 04:03 AM
Plato
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.
• December 6th 2007, 05:25 AM
kalagota
Quote:

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.