Negation of an implication (1 Viewer)

[yanujw]

What is a quick way to find the negation of implication and also how does this extend to the negation of an iff statement?

For example, if I wanted to find the negation of

How would I go about doing that, if the arrow was one-way or two-ways?

 

[5uckerberg]

Negation is just the opposite of what is shown here and using your example the negation of is .
Doing that then you can say is not a point inside the circle
In this case the original statement is true and the negation is also true.

 

[cossine]

What is a quick way to find the negation of implication and also how does this extend to the negation of an iff statement?

For example, if I wanted to find the negation of
View attachment 34965
How would I go about doing that, if the arrow was one-way or two-ways?

So it is important to know your definitions. Do you know how an if-then statement is written logically. A iff (if and only if) statement is just

(p -> q) and (p<-p=q).

You need

a) know how to write using p->q using logic then use Demorgan theorem to get the negation.

 

[yanujw]

Would this be correct working?

Define the following conditions as follows;
P: x^2 + y^2 < 1
~P: x^2 + y^2 >= 1
Q: (x,y) is a point inside the unit circle
~Q: (x,y) is a point outside the unit circle

Statement = P <-> Q
= (P -> Q) and (Q -> P)
= (Q or ~P) and (P or ~Q)
Negation = (P and ~Q) or (Q and ~P)

So the negation is that either P is true and Q is not, or Q is true and P is not, which is false.

In this case the original statement is true and the negation is also true.

Also, @5uckerberg isn't a statement and its negation logically opposite? So one is always true and the other is always false?

 

[cossine]

Would this be correct working?

Define the following conditions as follows;
P: x^2 + y^2 < 1
~P: x^2 + y^2 >= 1
Q: (x,y) is a point inside the unit circle
~Q: (x,y) is a point outside the unit circle

Statement = P <-> Q
= (P -> Q) and (Q -> P)
= (Q or ~P) and (P or ~Q)
Negation = (P and ~Q) or (Q and ~P)

So the negation is that either P is true and Q is not, or Q is true and P is not, which is false.



Also, @5uckerberg isn't a statement and its negation logically opposite? So one is always true and the other is always false?

Yes that is correct.

Good job