The assertion : A implies non(A), is it false?

Question

I have proved that $(1) \; A \to B.$

But I want prove that:

$(2)\; B \to \neg A.$

From (2), is (2) a false assertion? Because if (2) is true, this means that:

$A \to \neg A = \lnot A \lor \lnot A \equiv \lnot A$ is true.

Answer 1

So we start with the proven implication:

\begin{align*} &A \Rightarrow B \\ \end{align*}

We then consider the new implications:

\begin{align*} B \Rightarrow !A\\ A \Rightarrow !A \\ \end{align*}

This proves that A has to be False, because False statements can imply True statements, but True statements cannot imply False statements.

Answer 2

From (2), is (2) a false assertion? Because if (2) is true, this means that:

A→¬A=¬A∨¬A≡¬A is true.

Why would proving $\lnot A$ be a contradiction? Were you told that $A$ had to be true?

....

If not, then proving $A \to B$ and $B \to \lnot A$ will prove that $A$ is false. Which isn't a contradiction.

Notice $P \to Q$ and $P \to \lnot Q$ is not a contradiction. The can both be true but they are both true if and only if $P$ is false.

Likewise $P \to \lnot P$ is also not a contradiction. It will be true if and only $P$ is false.

So $A \to B$ and $B \to \lnot A$ will have proven $A \to \lnot A$ which isn't a contradiction if $A$ is false. However if you prove that both $A \to B$ and $B \to \lnot A$, that that is a proof that $A$ is false

The following may be useful:

$P \to \lnot P \iff \lnot P$

$P \to Q \iff \lnot Q \to \lnot P$

$(P\to Q)\land (P\to \lnot Q) \iff \lnot Q$

$(P \to Q) \land (\lnot P \to Q)\iff Q$

$(P \to Q) \land (Q \to \lnot P) \iff \lnot P$.

etc.

Also if $Q$ is true then $X \to Q$ is always true.

And if $P$ is false then $P \to X$ is always true.

So for any $X$ then $A \to X$ and $X\to \lnot A$ will be consistent and will both be true whenever $A$ is false (but if $A$ is true then one will be true and the other false depending upon whether $X$ is true or false.)