Is $A \to \neg A$ logically satisfiable if $A$ is false? 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. 
 A: $A\rightarrow B$ 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.
A: 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]
A: 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....
