Fix typo: tales -> takes

pdf/grandpa.tex

@@ -663,7 +663,7 @@ If we switch to $1$ when all node have already decided $0$, then we decide $0$.
 We claim that some configuration in the run $r$, where there are two runs from it where $A$ is always $1$ that decide $0$ and $1$. We call such states $1$-bivalent.
 To see this, assume for a contradiction that $r$ contains no such configurations. Then there are successive configurations $C$,$C'$ such that if $A$ return $1$ in the future from $C$ then we always decide $0$ but from $C'$, we always decide $1$.
 Let events be $(p,m,x)$ where node (processor/voter) $p$ receives message $m$ (which may be null) and executes some code where any calls to A return $x$ in $\{0,1\}$, then sends some messages. 
-Then there is some event $(p,m,0)$ that when applied to $C$ gives $C'$. Now suppose that $p$ goes offline at $C$, then if $A$ always returns $1$ afterwards, then we still decide $1$. Thus there is a run $r'$ that starts at $C$ where $p$ tales no steps, $A$ always returns $1$ and all other nodes still output $1$.
+Then there is some event $(p,m,0)$ that when applied to $C$ gives $C'$. Now suppose that $p$ goes offline at $C$, then if $A$ always returns $1$ afterwards, then we still decide $1$. Thus there is a run $r'$ that starts at $C$ where $p$ takes no steps, $A$ always returns $1$ and all other nodes still output $1$.
 But since $p$ takes no steps in $r'$, we can apply $r'$ after $(p, m, 0)$ and so we have that $C'$ has a run where $A$ always returns $1$ but decides $1$, which is a contradiction.
 
 Now let $C$ be a $1$-bivalent configuration. We can follow the FLP proof to show that there is a run from $C$ for which $A$ always returns $1$, all messages are delivered but all configurations are 1-bivalent and so the protocol never decides. This completes the proof by contradiction that there is no correct protocol.