Boolean operations AND

Question

When the $\times$ sign means AND, the possible results are

1. 0 × 0 = 0
2. 0 × 1 = 0
3. 1 × 0 = 0
4. 1 × 1 = 1

In other words, the result is 1 only if both the left operand AND the right operand are 1.

This is axiom?

Answer 1

A better way to explain what an axiom is:

An axiom is a statement that is treated as true without proof.

These are the assumed premises that you start with to make logical conclusions. What you have there is not an axiom. It's an operation $\times$ which you have defined as the construct AND.

The reason it's not an axiom is that we are restricting an action. In this case we are setting $\times$ as a valid operation within the Boolean algebraic structure we've built. This way if someone tries to use $\times$ they won't come up and say $0\times 0=1$.

By the way you can define $\times$ such that $0\times 0=1$ would be true.

Answer 2

What you've written (the truth-table for the AND-connective, essentially) and your summary is correct, but it expresses the definition of AND, but it is not an axiom. The four assignments you show define precisely when $a\times b$ (and in this case $a\land b$) is true, and when it is false.

Similarly, we define the inclusive OR as follows

1. 1+1 = 1

2. 1+0 = 1

3. 0+1 = 1

4. 0+0 = 0

Whenever at least one of the disjuncts is 1, then the disjunction is 1.

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The Exclusive Or: a XOR b is defined such that when one and only one of a, b is true, then a XOR B = 1.

This gives us:

1 XOR 1 = 0

1 XOR 0 = 1

0 XOR 1 = 1

0 XOR 0 = 0