1. T ⊃ G 2. S ⊃ G ∴ (T ∨ S) ⊃ G (T ⊃ G) & (S ⊃ G) given (~T ∨ G) & (~S ∨ G) by P ⊃ Q <=> ~P ∨ Q (~T & ~S) ∨ G by ∨ is distributive over & ~(T ∨ S) ∨ G by deMorgan's law. (T ∨ S) ⊃ G by P ⊃ Q <=> ~P ∨ Q Edwin