Thread: Standard logical equivalences

Hi all. i've got a one question well i need to show that p -> (q ^ r) and (p -> q) ^ (p -> r) are logically equivalent.

This is what i did ;

let p be A and B be (q^r)
A->B
not(A or B) - implication laws
not p or (q^r)
(not p or q) ^ (not p or q) - De Morgan laws
(not p or q) <=> p->q - implication laws
(not p or r) <=> p->r - implication laws

(p -> q) ^ (p -> r) <=> p -> (q ^ r)

Please anyone let me know if this is all right.

Regards

Originally Posted by Snowboarder

Hi all. i've got a one question well i need to show that p -> (q ^ r) and (p -> q) ^ (p -> r) are logically equivalent.

This is what i did ;

let p be A and B be (q^r)
A->B
not(A or B) - implication laws
not p or (q^r)
(not p or q) ^ (not p or q) - De Morgan laws
(not p or q) <=> p->q - implication laws
(not p or r) <=> p->r - implication laws

(p -> q) ^ (p -> r) <=> p -> (q ^ r)

Please anyone let me know if this is all right.

Regards

I think your basic idea is right, although the way you write it up seems unnecessarily messy to me. In particular I find it difficult to understand what introducing and gives you. Also, you probably meant to write that is equivalent to instead of . And I don't see an application of De Morgan's laws either...

Why not write the following:

Hello, Snowboarder!

Show that these are equivalent: .

DeMorgan does not apply here . . .



. .