Show (p → q) ∨ (p → r) is equivalent to p → (q ∨ r)

[bonfire09]

Homework Statement

(b) Show that (p → q) ∨ (p→ r) is equivalent to p → (q ∨ r).

Homework Equations

the ~ means negate

The Attempt at a Solution

Im not sure if i did this correctly
(p → q) ∨ (P → r)
(~p∨q) ∨ (~p∨r) used the conditional law p→q equivalent to ~p∨q
((~p∨q)∨~p)∨((~p∨q)∨r)) distributive law
(~p∨q)∨(~p∨q)∨r
(~p∨q)∨r
~p∨(r∨q) associative law
p →(q∨r)

 

[tiny-tim]

hi bonfire09!

all your steps are correct

however, after …

(p → q) ∨ (P → r)
(~p∨q) ∨ (~p∨r)

… don't you notice that they're all ∨ ,

so you can rearrange them (using the …?… law), and then use ~p∨~p = ~p

 

[HallsofIvy]

Are you required to do it that way? Setting up an 8 case "truth table" shows that both statements are false in case p= T, q= r= F, and true in all other cases.

 

[bonfire09]

That I am not sure upon. In Velleman's book he is not so clear about what he wants us to show. But I think what I did suffices.