1. ## 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

2. 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 $A$ and $B$ gives you. Also, you probably meant to write that $A\rightarrow B$ is equivalent to $\neg A\vee B$ instead of $\neg (A\vee B)$. And I don't see an application of De Morgan's laws either...

Why not write the following:
$\begin{array}{lcll}
p\rightarrow (q\wedge r) &\Leftrightarrow& \neg p \vee (q\wedge r) &\text{(implication law)}\\
&\Leftrightarrow& (\neg p \vee q)\wedge (\neg p\vee r)& \text{(distributivity law)}\\
&\Leftrightarrow& (p\rightarrow q)\wedge (p\rightarrow r)&\text{(implication law)}\end{array}$

3. Hello, Snowboarder!

Show that these are equivalent: . $
p \to (q \wedge r)\,\text{ and }\,(p \to q) \wedge (p \to r)$
DeMorgan does not apply here . . .

. . $\begin{array}{ccc}p \to (q \wedge r) && \text{Given} \\ \\ \sim p \vee (q \wedge r) && \text{Implication Law} \\ \\ (\sim p \vee q) \wedge (\sim p \vee q) && \text{Distributive Law} \\ \\ (p \to q) \wedge (p \to r) && \text{Implication Law }\end{array}$