~(p ► q)

**kathrynmath** · October 29th 2008, 08:22 PM

If p = My dog swims and q = the water is cold, what is ~(p ► q)? is it my dog swims if the water isn't cold?

**Soroban** · October 29th 2008, 08:51 PM

Hello, Kathryn!

$\begin{array}{ccc}p &=& \text{my dog swims} \\ q &=& \text{the water is cold}\end{array}$

What is: . $\sim (p \to q)$ ?

We know that: . $p \to q \;\;\Longleftrightarrow\;\;\sim\!p \vee q$

Then: . $\sim(p \to q)\;\;\Longleftrightarrow\;\;\sim(\sim\!p \vee q) \;\;\Longleftrightarrow\;\;p \:\wedge \sim\! q$ . DeMorgan's Law

In English: .My dog swims and the water is not cold.

Thread: ~(p ► q)

LinkBack

Thread Tools

Display

~(p ► q)

Search Tags