Is $A \to \neg A$ logically satisfiable if $A$ is false?

Question

Is it true that the implication $A \to \neg A$ is logically satisfiable if $A$ is false?

I beg your pardon if it is trivial but my logic is rusty.

Answer 1

$A\rightarrow B$ is equivalent to $\neg A\lor B$ by definition, so $A\rightarrow \neg A\iff \neg A\lor \neg A\iff \neg A,$ which is true if and only if $A$ is false.

Answer 2

Yes it is true that $$A \Longrightarrow \neg A$$ is satisfiable if $A$ is false. In fact, this is know as "vacuously true". We say a if-then statement, say $A \Longrightarrow B$, is false only when $A$ is true and $B$ is false. However, this question has already been asked multiple times: “false implies true” is a true statement [duplicate]

Answer 3

The truth table for $P \to Q$ is

$P =TRUE; Q=TRUE$ then $P\to Q = TRUE$

$P = TRUE; Q=FALSE$ then $P\to Q = FALSE$

$P = FALSE; Q=TRUE$ then $P\to Q=TRUE$

$P = FALSE; Q=FALSE$ then $P\to Q = TRUE$.

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So if $A$ is false then ... $A=FALSE$ and $\lnot A = TRUE$ and $A\to \lnot A = TRUE$.

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A good rule of thumb is:

If $P$ is false then

$P\to X$ is always true no matter what $X$ is. So if $A$ is false than $A\to X$ is always true, even if $X = \lnot A$.

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Another rule of thumb is: If $Q$ is true then

$X \to Q$ is always true no matter what $X$ is. So if $A$ is false then $\lnot A$ is true.

So $X \to \lnot A$ is always true no matter what $X$ is; even if $X = A$.

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Another rule of thumb (although not directly applicable here) is

The only way for $P \to Q$ to be false is if $P$ is true and $Q$ is false. Any thing else... then $P \to Q$ is true.

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It may seem weird but when you get used to it

$P\to Q \equiv (\lnot P)\lor Q$

So so $A\to \lnot A$ is true if either $\lnot A$ is true or..... $\lnot A$ is true....