Question

Using the laws of propositional equivalence,  show that  p → p ∧ ( p → q) is logically equivalent to p → q. That means a chain of equivalences starting with the first expression and ending with p → q. Remember: The connective ∧ has higher precedence than the connective → so be sure to parse the original expression correctly!

To start you off here's a possible first step. p → p ^ ( p → q) ≡   p → p ^ ( ¬p ∨ q) .

Type or paste your answer into the Write Submission box.  You may use ~ in place of ¬ if you're typing. Similarly, you may use --> in place of →.  

Each step must use exactly one law, except in the case of commutativity and associativity which you may combine with other laws. For example, a step might involve replacing ¬p ∨ p by true using the law A ∨ ¬A ≡ true along with commutativity. Or you could use two steps if you prefer: replace ¬p ∨ p by p ∨ ¬p, then replace the latter by true .  

Answer 1

given p → p ^ ( p → q)

p → p ^ ( p → q) ≡   p → p ^ ( ¬p ∨ q) using implication law

p → p ^ ( p → q) ≡  p → ((p ^ ~p) v (p^q)) using distributive law

p → p ^ ( p → q) ≡   p → (F v (p^q)) using contradiction

p → p ^ ( p → q) ≡ p → (p^q) using domination law

   p → p ^ ( p → q) ≡ (~p v (p^q)) using implication law

  p → p ^ ( p → q) ≡ (~p v p)^(~p v q)) using distributive law

  p → p ^ ( p → q) ≡ T ^ (~p v q) using tautology

p → p ^ ( p → q) ≡ ~p v q using identity law

p → p ^ ( p → q) ≡ p → q using implication law

hence proved