Is Logical Equivalence of Conditional Statements a valid title for this content?

  • Thread starter Thread starter bonfire09
  • Start date Start date
  • Tags Tags
    Equivalence
bonfire09
Messages
247
Reaction score
0

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)
 
Physics news on Phys.org
hi bonfire09! :smile:

all your steps are correct

however, after …
bonfire09 said:
(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 :wink:
 
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.
 
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top