The Modus Ponens states that (p ∧ (p ⇒ q)) ⇒ q. Prove this using Boolean laws.

To prove: $(p \wedge (p \Rightarrow q)) \Rightarrow q \equiv 1$ (a tautology) 1. $p \Rightarrow q = p' + q$ (Conditional/Implication law) 2. $p \wedge (p\Rightarrow q) = p\cdot(p'+q)$ (substituting step 1) 3. $= p\cdot p' + p\cdot q$ (Distributive law) 4. $= 0 + p\cdot q$ (Complement law: $p\cdot p'=0$) 5. $= p\cdot q$ (Identity law) 6. $(p\wedge(p\Rightarrow q))\Rightarrow q = (p\cdot q)\Rightarrow q = (p\cdot q)'+q$ (Conditional law, using step 5) 7. $= p'+q'+q$ (De Morgan's law) 8. $= p'+(q'+q)$ (Associative law) 9. $= p'+1$ (Complement law: $q'+q=1$) 10. $= 1$ (Null/Dominance law) Hence $(p\wedge(p\Rightarrow q))\Rightarrow q = 1$ for all values of $p,q$ — it is a tautology, which proves Modus Ponens. (Verified by truth table.)