Confusion with Implication

Question

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
  

I am missing something or have some conceptual flaws but unable to find, Please help me

Thank You

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

Answer 1

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!!

Answer 2

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.

Answer 3

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.

Answer 4

"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.$$A\to B ~\iff ~ \neg A\vee B$$