**Proof.**
The equivalence of (3) and (4) is More on Algebra, Lemma 15.33.5.

If $A \to B$ is syntomic, then we can find a cocartesian diagram

\[ \xymatrix{ B_0 \ar[r] & B \\ A_0 \ar[r] \ar[u] & A \ar[u] } \]

such that $A_0 \to B_0$ is syntomic and $A_0$ is Noetherian, see Algebra, Lemmas 10.127.18 and 10.168.9. By Lemma 23.11.4 we see that $\mathop{N\! L}\nolimits _{B_0/A_0}$ is perfect of tor-amplitude in $[-1, 0]$. By More on Algebra, Lemma 15.85.3 we conclude the same thing is true for $\mathop{N\! L}\nolimits _{B/A} = \mathop{N\! L}\nolimits _{B_0/A_0} \otimes _{B_0}^\mathbf {L} B$ (see also More on Algebra, Lemmas 15.66.13 and 15.74.9). This proves that (3) implies (2).

Assume (1). By More on Algebra, Lemma 15.85.3 for every ring map $A \to k$ where $k$ is a field, we see that $\mathop{N\! L}\nolimits _{B \otimes _ A k/k}$ has tor-amplitude in $[-1, 0]$ (see More on Algebra, Lemma 15.66.13). Hence by Lemma 23.11.4 we see that $k \to B \otimes _ A k$ is a local complete intersection homomorphism. Thus $A \to B$ is syntomic by definition. This proves (1) implies (3) and finishes the proof.
$\square$

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