# Confusion with Implication

Forgive me if its a dumb question, I just started reading Mathematical Logic

Question:-

Let we have an Implication,

A ⟹ B

And its Truth Table is

This Implication is true for All False values of “A” irrespective the value of B.

By this we can conclude that "If not A then B is false/true" both are true. So we can say "If not A then B" because B only has two value True or False and here B can be any thing.

By this reasoning "If A then B" should be equal to If not A then B

But when we write a mathematical formula and calculate their truth table then both are different

• "If A then B"

Mathematical Formula :-

A ⟹ B

Truth Table:-

Here I am taking only one case(i.e. B is True)

• "If not A then B"

Mathematical Formula :-

not A ⟹ B

A    B             If not A then B

T    T                    T
T    F                    T
F    T                    T
F    F                    F


Thank You

P.S. :- Sorry in advance because my English is not upto that mark. Edits are welcome :)

• @amWhy I know that my question is different I guess – Bhaskar Feb 19 '18 at 17:29
• When A is true and B is false you get false in implication. But replacing A by not A does not work in this case: not A is false and B is false we get true. – Shahab Feb 19 '18 at 17:31
• @Bhaskar: Remember not A is not always false. So we cannot say "If not A then B" is always true as you seem to say. – Shahab Feb 19 '18 at 17:35
• @Shahab I am asking what is wrong with my reasoning. I know "replacing A by not A does not work". I want an Intuitive Explanation not mathematical formula based :) – Bhaskar Feb 19 '18 at 17:38
• Ok...sorry if I misinterpreted. What do you mean when you say, : "By this we can conclude that "If not A then B is false/true" both are true. " What does both mean? – Shahab Feb 19 '18 at 17:52

First of all, I am confused why you would say that $\neg A \rightarrow B$ should be equivalent to $A \rightarrow B$, given that you just argued that the value of $B$ should not matter, rather than the value of $A$! In fact, when $A$ is true it is no longer the case that the value of $B$ doesn;t matter, and so you immediately get that $\neg A \rightarrow B$ is not the same as $A \rightarrow B$.

What would have made a little more sense is if you would have focused on $A \rightarrow \neg B$ instead, because (as you correctly observed) if $A$ is false, then $A \rightarrow B$ has the same truth-value as $A \rightarrow \neg B$ (namely True). However, that still does not mean that they are equivalent, because equivalence means that they should have the same truth-value under any conditions (and again, you have only shown them to have the same truth-value under the condition that $A$ is False). And so $A \rightarrow \neg B$ is also not equivalent to $A \rightarrow B$

Finally, if you are trying to change the $A$ into a $\neg A$, because $A \rightarrow B$ is true when $A$ is false ... well, that makes even less logical sense. Here is an example to demonstrate your faulty logic. Take statement $\neg A$. This statement is true when $A$ is false ... ok, so by your logic we should be able to change the $A$ with a $\neg A$ and get the same statement? No, because changing the $A$ with a $\neg A$ in $\neg A$ gives us $\neg \neg A$, which is equivalent to just $A$ ... which is not at all equivalent to the original $\neg A$.

Don't confuse statements with their truth-values!!

• Please forgive me, I know it's a very stupid question ........ What is the difference between "If not A then B" and putting the value "Not A" in "If A then B". ........... I know they are different but unable to get it intuitively – Bhaskar Feb 19 '18 at 19:39
• @Bhaskar I am not sure what you mean by 'putting the value of $neg A$ in $A \rightarrow B$' ... you can replace parts of a statement with a different statement, sure, but that usually gives you a statement with different truth-conditions, as I show in my answer. If you mean that you can substitute $\neg A$ for $A$ because $A$ is false ... absolutely not! Again, it changes the statement. Think about this: I have statement $A$. Now you tell me $A$ is false .. ok, so I should be able to change the $A$ into $\neg A$? No, because when $A$ is false, $\neg A$ is True! – Bram28 Feb 19 '18 at 19:51
• But isn't"A" is false and "not A" are same I mean have same meaning – Bhaskar Feb 19 '18 at 19:59
• @Bhaskar Yes ... but you are confusing the statement with its truth-value. A statement is just that: a statement. And the statement can be true or false. Just because a statement $A$ is false does not mean you can substitute $\neg A$ for $A$ in some other statement. When you do, you change the statement – Bram28 Feb 19 '18 at 20:04
• Yes you are right I am confusing a statement with its truth value.... But I don't know why I am still not confident that I got the point.... Can please elaborate... Its a request – Bhaskar Feb 19 '18 at 20:13

Here's part of your problem. You wrote, "If not A then B is false/true". Well, you then analyze "If not A then B". But when you wrote "false/true", you really meant "false OR true". But if you write "If not A then B", you're trying to show a much stronger conclusion than is warranted (A AND B is a stronger conclusion than A OR B), because you're going to need B true, period. As a matter of fact, the statement "If A then (true or false)" is always true, because the consequent is always true.

Another part of your problem, I think, is that you're trying to show these other formulas are true in the context of the statement "If A then B". But your later truth tables don't reflect that.

Near the beginning of your question, you noticed that, because of the truth table, if $A$ is false, then $A\to B$ is true, regardless of whether $B$ is true or false. That is correct, and it could be expressed by saying that $(\neg A)\to(A\to B)$ is always true. But then you seem to have inferred that $(\neg A)\to B$ and also that $(\neg A)\to(\neg B)$ are true, and I'm afraid there's no justification for either of those inferences.

"If A, then B" is clearly not equivalent to "if not A, then B".

However it is equivalent to "not A or B".   You may be thinking of that.

$$\begin{array}{c:c|c|c:c:c}A & B & \neg A & \neg A\vee B& A\to B ~~& \neg A \to B~~\\\hline T&T&F&T~~~~&T~~~~&T~~~~\\T&F&F&F~~\star&F~~\star&T~~~~\\F&T&T&T~~~~&T~~~~&T~~~~\\ F&F&T&T~~~~& T~~~~&F~~ \star\end{array}$$

$A\to B$ is only false when $A$ is true and $B$ is false.   As is $\neg A \vee B$.   They are equivalent

$\neg A\to B$ is false when $A$ is false and $B$ is false. (ie $\neg A$ is true and $B$ is false.)  So clearly that is not equivalent to the others.$$A\to B ~\iff ~ \neg A\vee B$$