- ~(p ∧ q) ⇔ ~p ∨ ~q
- ~(p ∨ q) ⇔ ~p ∧ ~q
- Show that this is true:
-
| p | q | p ∧ q | ~(p ∧ q) | ~p | ~q | ~p ∨ ~q |
|---|
| T | T | T | F | F | F | F |
| T | F | F | T | F | T | T |
| F | T | F | T | T | F | T |
| F | F | F | T | T | F | T |
- We can change this anyway we wish
- ~(~p ∧ q) ⇔ p ∨ ~q
- p ∨ ~q ⇔ ~(~p ∧ q)
- They ask us to use De Morgan's on statements:
-
Chase is not hiding or the pitcher is broken.
p: Chase is hiding
q: The pitcher is broken
statement: ~p ∨ q
Equivalent: ~(p ∧ ~q)
It is not true that Chase is hiding and the pitcher is not broken.
- We also have that p → q ⇔ ~p ∨ q
| p | q | p → q | ~p | ~p ∨ q |
|---|
| T | T | T | F | T |
| T | F | F | F | F |
| F | T | T | T | T |
| F | F | T | T | T |
-
If you study then you will get an "A".
p: you study
q: you get an "A"
p → q ⇔ ~p ∨ q
You do not study or you will get an "A".
- We can derive the negation
p → q ⇔ ~p ∨ q (We know this from above)
~(p → q) ⇔ ~(~p ∨ q) (negate both sides)
⇔ p ∧ ~q (apply De Morgans')
- It is not true that if you study then you will get an "A" is equivalent to You will study and you will not get an "A".