**Proof.**
The equivalence of (1) and (2) follows from the fact that $\mathfrak m_ AA^\wedge $ is the maximal ideal of $A^\wedge $ (and similarly for $B$) and faithful flatness of $B \to B^\wedge $. For example if $A^\wedge \to B^\wedge $ is unramified, then $\mathfrak m_ AB^\wedge = (\mathfrak m_ AB)B^\wedge = \mathfrak m_ BB^\wedge $ and hence $\mathfrak m_ AB = \mathfrak m_ B$.

Assume the equivalent conditions (1) and (2). By Lemma 41.3.3 we see that $A^\wedge \to B^\wedge $ is finite. Hence $A^\wedge \to B^\wedge $ is of finite presentation, and by Algebra, Lemma 10.151.7 we conclude that $A^\wedge \to B^\wedge $ is unramified at $\mathfrak m_{B^\wedge }$. Since $B^\wedge $ is local we conclude that $A^\wedge \to B^\wedge $ is unramified.

Assume (3). By Algebra, Lemma 10.151.5 we conclude that $A^\wedge \to B^\wedge $ is an unramified homomorphism of local rings, i.e., (2) holds.
$\square$

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