Show that
a. ~(pvq)v(~p^q) equivalent (~p^~q)v(~p^q)"De Morgan's law"
b. ~(pvq)v(~p^q) equivalent ~p^(~qvp) "Distributive law"
To show that the given expressions are equivalent, we can use truth tables to analyze all possible truth values of the statements involved.
a. De Morgan's Law:
We need to show that ~(p ∨ q) ∨ (~p ^ q) is equivalent to (~p ∧ ~q) ∨ (~p ^ q).
Truth Table for ~(p ∨ q) ∨ (~p ^ q):
| p | q | p ∨ q | ~p | ~(p ∨ q) | ~p ^ q | ~(p ∨ q) ∨ (~p ^ q) |
| T | T | T | F | F | F | F |
| T | F | T | F | F | F | F |
| F | T | T | T | F | T | T |
| F | F | F | T | T | F | T |
Truth Table for (~p ∧ ~q) ∨ (~p ^ q):
| p | q | ~p | ~q | ~p ∧ ~q | ~p ^ q | (~p ∧ ~q) ∨ (~p ^ q) |
| T | T | F | F | F | F | F |
| T | F | F | T | F | F | F |
| F | T | T | F | F | F | F |
| F | F | T | T | T | F | T |
As we can see from the truth tables, both expressions have the same truth values for all possible combinations of p and q. Therefore, ~(p ∨ q) ∨ (~p ^ q) is equivalent to (~p ∧ ~q) ∨ (~p ^ q) by De Morgan's Law.
b. Distributive Law:
We need to show that ~(p ∨ q) ∨ (~p ^ q) is equivalent to ~p ^ (~q ∨ p).
Truth Table for ~(p ∨ q) ∨ (~p ^ q):
| p | q | p ∨ q | ~p | ~(p ∨ q) | ~p ^ q | ~(p ∨ q) ∨ (~p ^ q) |
| T | T | T | F | F | F | F |
| T | F | T | F | F | F | F |
| F | T | T | T | F | T | T |
| F | F | F | T | T | F | T |
Truth Table for ~p ^ (~q ∨ p):
| p | q | ~p | ~q | ~q ∨ p | ~p ^ (~q ∨ p) |
| T | T | F | F | T | F |
| T | F | F | T | T | F |
| F | T | T | F | T | F |
| F | F | T | T | T | T |
As we can see from the truth tables, ~(p ∨ q) ∨ (~p ^ q) is not equivalent to ~p ^ (~q ∨ p) by the Distributive Law, as they do not have the same truth values for all possible combinations of p and q.