p ∧ (q ∨ r ) ≡ (p ∧ q) ∨ (p ∧ r )p ∨ (q ∧ r ) ≡ (p ∨ q) ∧ (p ∨ r )
To prove the first statement, we need to show that both sides of the equation are logically equivalent
To prove the first statement, we need to show that both sides of the equation are logically equivalent. We can do this by creating truth tables for both sides and showing that they have the same truth values under all possible combinations of truth and false values for the variables p, q, and r.
For p ∧ (q ∨ r ):
| p | q | r | q ∨ r | p ∧ (q ∨ r ) |
|:—:|:—:|:—:|:—-:|:———-:|
| T | T | T | T | T |
| T | T | F | T | T |
| T | F | T | T | T |
| T | F | F | F | F |
| F | T | T | T | F |
| F | T | F | T | F |
| F | F | T | T | F |
| F | F | F | F | F |
For (p ∧ q) ∨ (p ∧ r ):
| p | q | r | p ∧ q | p ∧ r | (p ∧ q) ∨ (p ∧ r ) |
|:—:|:—:|:—:|:—-:|:—-:|:—————-:|
| T | T | T | T | T | T |
| T | T | F | T | F | T |
| T | F | T | F | T | T |
| T | F | F | F | F | F |
| F | T | T | F | F | F |
| F | T | F | F | F | F |
| F | F | T | F | F | F |
| F | F | F | F | F | F |
As we can see from the truth tables, both sides of the equation have the same truth values for all possible combinations of truth and false values for p, q, and r. Therefore, we can conclude that p ∧ (q ∨ r ) ≡ (p ∧ q) ∨ (p ∧ r ).
To prove the second statement, we can follow the same process by creating truth tables for both sides of the equation.
For p ∨ (q ∧ r ):
| p | q | r | q ∧ r | p ∨ (q ∧ r ) |
|:—:|:—:|:—:|:—-:|:———-:|
| T | T | T | T | T |
| T | T | F | F | T |
| T | F | T | F | T |
| T | F | F | F | T |
| F | T | T | T | T |
| F | T | F | F | F |
| F | F | T | F | F |
| F | F | F | F | F |
For (p ∨ q) ∧ (p ∨ r ):
| p | q | r | p ∨ q | p ∨ r | (p ∨ q) ∧ (p ∨ r ) |
|:—:|:—:|:—:|:—-:|:—-:|:——————:|
| T | T | T | T | T | T |
| T | T | F | T | T | T |
| T | F | T | T | T | T |
| T | F | F | T | F | F |
| F | T | T | T | T | T |
| F | T | F | T | F | F |
| F | F | T | F | T | F |
| F | F | F | F | F | F |
Again, both sides of the equation have the same truth values for all possible combinations of truth and false values for p, q, and r. Therefore, we can conclude that p ∨ (q ∧ r ) ≡ (p ∨ q) ∧ (p ∨ r ).
More Answers:
[next_post_link]