This is a follow-up question to Does $|(X\times\{0\}) \cup (X\times\{1\})| \leq |X|$ for $X$ infinite imply ${\sf AC}$?Does $|(X\times\{0\}) \cup (X\times\{1\})| \leq |X|$ for $X$ infinite imply ${\sf AC}$?
Consider the statements
$(\text{S}1)$ For any infinite set $X$ there is an injection $\varphi$ from $(X\cup\{X\})$ into $X$.
and
$(\text{S}2)$ For any infinite set $X$ there is an injection $\varphi$ from $(X\times\{0\}) \cup (X\times\{1\})$ into $X$.
Is it true that ${\sf ZF} < {\sf ZF}+(\text{S1}) < {\sf ZF}+(\text{S}2)$? (Here $<$ means "strictly weaker".)
