MAT102 Lecture 4

$A \otimes B = \neg (A \land B)$
We will make a truth table to show how $\neg A$ is the same as $A \otimes A$ or $\neg (A \land A)$ .

$A$	$B$	$\neg A$	$\neg (A \land A)$ or $A \otimes A$
true	true	false	false
true	false	false	false
false	true	true	true
false	false	true	true

We see how not A and A is the exact same as not A in columns 3 and 4.

Here is another truth table which will show that $(A \otimes B) \otimes (A \otimes B) = A \land B$

$A$	$B$	$(A \otimes B)$	$(A \otimes B) \otimes (A \otimes B)$	$A \land B$
true	true	false	true	true
true	false	true	false	false
false	true	true	false	false
false	false	true	false	false

$(A \otimes B) \otimes (A \otimes B) = \neg (A \otimes B)$
$(A \otimes B) \otimes (A \otimes B) = \neg (\neg (A \land B))$
$(A \otimes B) \otimes (A \otimes B) = A \land B$

We can see that these are the same.
Through trial and error I find that:
- $A \lor B = (A \otimes A) \otimes (B \otimes B)$

$A$	$B$	$A \lor B$	$\neg (A \land A)$ or $A \otimes A$	$\neg (B \land B)$ or $B \otimes B$	$(A \otimes A) \otimes (B \otimes B)$
true	true	true	false	false	true
true	false	true	false	true	true
false	true	true	true	false	true
false	false	false	true	true	false

$\neg (A \lor B) = \neg A \land \neg B$
$A \lor B = \neg (\neg A \land \neg B) = (\neg A) \otimes (\neg B)$
$A \lor B = \neg (\neg A \land \neg B) = (A \otimes A) \otimes (B \otimes B)$