Took a bit of time to type up the explanation for exercise 3 properly, but here it is.
Let:
A be (x == y)
B be (x == z)
¬A be NOT A (x != y) (and similarly ¬B be NOT B (x != z))
A ∧ B be...
Type: Posts; User: jashburn
Re: problem with understanding boolean logic
Took a bit of time to type up the explanation for exercise 3 properly, but here it is.
Let:
A be (x == y)
B be (x == z)
¬A be NOT A (x != y) (and similarly ¬B be NOT B (x != z))
A ∧ B be...
Re: problem with understanding boolean logic
My boolean algrebra is rusty, but here goes...
For exercise 4, De Morgan's Theorem states that (using Java notation):
!(A && B) == !A || !B
If you substitute A with (x <= y), and B with (y...