From time to time, although not so commonly, I see $A \supset B$ written with seemingly the same meaning as $A \Rightarrow B$.
Ex. in Constructive Type Theory and Interactive Theorem Proving, on page 8.
A proof of $A \supset B$ is a method which transforms a proof of $A$ to a proof of $B$.
This suggest $A \supset B$ means the same as $A \Rightarrow B$, isn't it?
$A \Rightarrow B$ means: whenever $A$ holds, then $B$ holds. However, this also means $B$ may holds when $A$ do not. Which in turn means the set of the cases where $A$ holds is a subset of the set of the cases where $B$ holds. Is this OK? If that's OK, $A \Rightarrow B$ should be written as $A \subset B$ instead of $A \supset B$.
Why is this not so? Or am I wrong in some way? If I'm wrong, where am I so?
Edit: oops, duplicate of this one: Using $p\supset q$ instead of $p\implies q$ . Sorry.
The symbol $\supset$ (called "horseshoe") was used a century ago for the conditional connective "if-then"(see The Notation in Principia Mathematica , and after replaced (mainly) by $\rightarrow$ or $\Rightarrow$.
It does not mean that the sentence $A$ is "included" into the sentence $B$.
Of course, there is a "relation", through The Algebra of Logic Tradition and Giuseppe Peano.
Note
In the specific example you are referencing, the Author wants to differentiate the "if-then" connective from the (more heavy used in his presentation) symbol for "mapping" : $\rightarrow$.
Thus, he pulled out from the attic the symbol $\supset$.