1. ## logic notation

Hello,

I'm working on this assignment that is due in 2 hours, and I'm not quite sure about these two questions that I have yet to complete before I can submit this assignment.

1. Give the reasons for each of the steps provided to validate the argument

[(P → Q) /\ (¬R \/ S) /\ (P \/ R)] → (¬Q → S)

Step:

a) ¬(¬Q → S)
b) ¬Q /\ ¬S
c) ¬S
d) ¬R \/ S
e) ¬R
f) P → Q
g) ¬Q
h) ¬P
i) P \/ R
j) R
k) ¬R /\ R
l) therefore ¬Q → S

2. Find the value of the Boolean expression

(wz + yx' + wy')(y'x + zx') + ( (w + y)(x' + y) )'

if

a) w = x = 0 and y = z = 1
b) w = y = 1 and x = z = 0

I shall review this material after I submit the assignment and will learn and understand it better. Right now, I just really need to get this handed in. Any help would be much appreciated.

Thank you kindly!

2. Hello, spider_dude!

1. Give the reasons for each of the steps provided to validate the argument

$\bigg[(P \to Q) \wedge (\sim\! R \vee S) \wedge (P \vee R)\bigg] \to (\sim\! Q \to S)$

$\begin{array}{ccc}
(P \to Q) \wedge (S\:\vee \sim\! R) \wedge (R \vee P) & & \vee\text{ is commutative} \\ \\
(P\to Q) \wedge (\sim\! S \to \sim\! R) \wedge ( \sim R\! \to P) & & \text{de{f
}. of implication} \\ \\
(\sim\! S \to\sim\! R) \wedge (\sim\! R\to P) \wedge (P \to Q) & & \wedge\text{ is assoc. \& comm.} \\ \\
\sim\!S \to Q & & \text{syllogism} \\ \\
\sim\!Q\to S & & \text{contrapositive}
\end{array}$