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$S1$ is equivalent to "every infinite set is Dedekind-infinite", so $\sf ZF<ZF+(S1)$, because it is consistent that infinite Dedekind-finite sets exist.

$\sf ZF+(S1)<ZF+(S2)$ follows using results from this answerthis answer, which implies that $\sf DC_\kappa$ doesn't imply $S2$, and hence nonexistence of infinite Dedekind-finite sets, which is a corollary to $\sf DC(=DC_\omega)$.

$S1$ is equivalent to "every infinite set is Dedekind-infinite", so $\sf ZF<ZF+(S1)$, because it is consistent that infinite Dedekind-finite sets exist.

$\sf ZF+(S1)<ZF+(S2)$ follows using results from this answer, which implies that $\sf DC_\kappa$ doesn't imply $S2$, and hence nonexistence of infinite Dedekind-finite sets, which is a corollary to $\sf DC(=DC_\omega)$.

$S1$ is equivalent to "every infinite set is Dedekind-infinite", so $\sf ZF<ZF+(S1)$, because it is consistent that infinite Dedekind-finite sets exist.

$\sf ZF+(S1)<ZF+(S2)$ follows using results from this answer, which implies that $\sf DC_\kappa$ doesn't imply $S2$, and hence nonexistence of infinite Dedekind-finite sets, which is a corollary to $\sf DC(=DC_\omega)$.

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Wojowu
Wojowu
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$S1$ is equivalent to "every infinite set is Dedekind-infinite", so $\sf ZF<ZF+(S1)$, because it is consistent that infinite Dedekind-finite sets exist.

$\sf ZF+(S1)<ZF+(S2)$ follows using results from this answer, which implies that $\sf DC_\kappa$ doesn't imply $S2$, and hence nonexistence of infinite Dedekind-finite sets, which is a corollary to $\sf DC(=DC_\omega)$.