There are two ways to do this same proof. In this first one, I've disallowed the derived rules (including De Morgan's and Modus Tollens) so you need to prove the goal using ~I and V. You need to do the vE inside the ~| subproof, so this is our first example of nested subproofs. 1 10 pts -A, B AF -(A v B) 1| not A :PR 2 B > A:PR 3|| A or B :AS 4|||A :AS 5||| 6||| _|_ :-E?,? 7|| not A :R1 8||| B :AS 9||| 10||| __ :-E?,? 11|| _I_ :VE3, 4-?,?-? 12] not (A or B) :-13-? A Submit