2. (8 marks, 4 each) Using only the four rules MP, CP, DN and R and the following definitions 1. p =df p 2. p q =df p q 3. p & q =df (p q) Prove DeM: (p & q) p q and DL: p q, p r, q r r Use definitions as you would equivalence rules and list the reason as DF and the number of the definition, so DF1 for the first negation definition, etc. As an example, here is a proof of MT. I use small letters as variables to represent arbitrary statements of SL. MT: p q, q p 1. p q Prem 2. q Prem 3. p Supp/CP 4. q 1,3 MP 5. q 2 DF1 6. 5,4 MP 7. p 3-6 CP 8. p 7 DF1 DeM: (p & q) p q DL: p q, p r, q r r