Formal Proof
The laws may be proven directly using truth tables; "1" represents true, "0" represents false.
First we prove: ¬(p ∨ q) ⇔ (¬p) ∧ (¬q).
p | q | p ∨ q | ¬(p ∨ q) | ¬p | ¬q | (¬p) ∧ (¬q) |
---|---|---|---|---|---|---|
0 | 0 | 0 | 1 | 1 | 1 | 1 |
0 | 1 | 1 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 0 | 0 | 1 | 0 |
1 | 1 | 1 | 0 | 0 | 0 | 0 |
Since the values in the 4th and last columns are the same for all rows (which cover all possible truth value assignments to the variables), we can conclude that the two expressions are logically equivalent.
Now we prove ¬(p ∧ q) ⇔ (¬p) ∨ (¬q) by the same method:
p | q | p ∧ q | ¬(p ∧ q) | ¬p | ¬q | (¬p) ∨ (¬q) |
---|---|---|---|---|---|---|
0 | 0 | 0 | 1 | 1 | 1 | 1 |
0 | 1 | 0 | 1 | 1 | 0 | 1 |
1 | 0 | 0 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 0 | 0 | 0 | 0 |
Read more about this topic: De Morgan's Laws
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