Why is it that when A is false and B is false, we infer that A->B is true?

Why is false and false false?

If the left side of the expression is “falsey”, the expression will return the left side. If the left side of the expression is “truthy”, the expression will return the right side. That’s it. So in false && false , the left side is “falsey”, so the expression returns the left side, false .

Why is implication not A or B?

For instance, logical implication: A implies B if whenever A is true, B is true too. It’s usually interpreted to mean (see discussion in Section 14.2) that this can only be false when A is true and B is false, so an equivalent proposition is “B or not A”.

Does A and B imply a?

⇔ ((A implies B) and (B implies A)). In other words, A and B are equivalent exactly when both A ⇒ B and its converse are true. (A implies B) ⇔ (¬B implies ¬A).

What does it mean if A implies B?

“A implies B” means that B is at least as true as A, that is, the truth value of B is greater than or equal to the truth value of A. Now, the truth value of a true statement is 1, and the truth value of a false statement is 0; there are no negative truth values.

Why is false or false true?

As you can see from the truth table, it is only if both conditions are true that the conjunction will equate to true. If one or other or both of the conditions in the conjunction are false, then the conjunction equates to false.
AND truth table.

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Why True implies false is false?

In classical propositional logic, an implication is true if it is the case that if is true, then must be true as well. So when is false, we do not care about the truth value of ; the implication is still true. Now consider the implication where is falsehood and is arbitrary.

Is A or B equal to not A and not B?

The negation of a conjunction (logical AND) of 2 statements is logically equivalent to the disjunction (logical OR) of each statement’s negation. That sounds like a mouthful, but what it means is that “not (A and B)” is logically equivalent to “not A or not B”.

Is a B equivalent to a B?

Another way to eyeball it is to see that A+B is almost like A|B except for the carry in the bottom row. A&B isolates that bottom row for us, A-A&B moves those isolated cased up two rows in the + table, and the (A-A&B)+B becomes equivalent to A|B.