Are infinite Dedekind-finite cardinals “cancellable” in $\sf ZF$? Let $\mathfrak p$ be infinite Dedekind-finite. It is known that, in $\sf ZF$, for any non-zero finite ordinal $n$, we have $|n \cdot |A|| = |n \cdot |B|| \implies |A| = |B|$ for any sets $A$ and $B$. Motivated by this, is it true that $\mathfrak p \cdot |A| = \mathfrak p \cdot |B| \implies |A| = |B|$ for any sets $A$ and $B$, where $\mathfrak p$ is arbitrary and as above? Or at least is there a model of $\sf ZF$ with a Dedekind-finite set in it which satisfies this property?
Note that if $A$ and $B$ are finite, then this holds, as (Dedekind-)finite union of Dedekind-finite sets is Dedekind-finite.
This question arised because of the fact that, in $\sf ZFC$, all “cancellable” sets are finite, this made me wonder if there are sets in $\sf ZF$ which are “cancellable” but not finite, or at least if there is a spesific set in a spesific model of $\sf ZF$ with this property, thank you.
 A: (This is a partial answer), I am not sure about arbitrary sets $A,B$, but I can give the following $2$ points:

*

*If $\mathfrak p$ is a non-zero Dedekind finite cardinal, and $A,B$ are well orderable, then $\mathfrak p \cdot |A| = \mathfrak p \cdot |B| \implies |A| = |B|$.


*It is consistent that $\mathfrak p \cdot \mathfrak n = \mathfrak p \cdot \mathfrak m \implies \mathfrak n = \mathfrak m$ for all non-zero Dedekind finite $\mathfrak{p,n,m}$.

Proof of (1):
Let $\mathfrak p=|P|$ be non-zero Dedekind finite and $|A|<|B|$ aleph numbers, we will show that $\mathfrak p \cdot |A| < \mathfrak p \cdot |B|$.
It is clear that $\mathfrak p \cdot |A| ≤ \mathfrak p \cdot |B|$, assume that they are equal and let $f:A×P→B×P$ be a bijection, we will recursively define an injection from $ω$ to $P$.
Assume that $g_k:k\to P$ is injective, define on $B\times f_k''k$ the following (modified) lexicographic order: $(a,g_k(i))<'(b,g_k(j))⇔i<j∨(i=j∧a<b)$.
This is a well ordering, and because $g_k''k$ is finite, $|A\times g_k''k|<'|B\times g_k''k|$.
Let $(a,g_k(i))$ be the $<'$-first element such that $f((a,g_k(i)))=(b,p)∈P×B\setminus P×g_k''k$, and define $g_{k+1}:k+1→P$ as $g_k\cup\{(k, p)\}$.
Then $\bigcup_{i\in\omega}g_i$ is an injection from $ω$ to $P$, contradiction.

"Proof" of (2):
First note that it is consistent that the class of Dedekind finite cardinals is linearly ordered (as is done in this paper by Gershon Sageev, mentioned by Noah).
Then we take the result of Ellentuck that if the class of Dedekind finite cardinals is linearly ordered, then that class form a model of Peano Axioms.
This theorem supposedly appears in his paper The Universal Properties of Dedekind Finite Cardinals
 But I couldn't find it there, as well as in his book The Theory of Dedekind Finite Cardinals
 that I couldn't find anywhere.
I did find the proof of the theorem in his paper A Model of Arithmetic, but I couldn't find any Legal prints of this paper.

Addon: As @user14111 correctly said, we don't actually need a full model of Peano Axioms, we just need to note the following fact


*If $\mathfrak{p,n,m}$ are Dedekind finite such that $\mathfrak n < \mathfrak m$, then $\mathfrak p\cdot \mathfrak n < \mathfrak p \cdot \mathfrak m$
Then if the Dedekind finite cardinals are linearly ordered, we get $\mathfrak p \cdot \mathfrak n = \mathfrak p \cdot \mathfrak m \implies \mathfrak n = \mathfrak m$.
To prove (3), let's note (given $P,N,M$ sets of those cardinality) that $P×N⊊P×M$ and that product of Dedekind finite cardinals is Dedekind finite, hence we get $\mathfrak p\cdot \mathfrak n < \mathfrak p \cdot \mathfrak m$.

Addon 2: After reading a little bit more of "The Universal Properties of Dedekind Finite Cardinals", it appears to be that Ellentuck actually proved that with ACF (families of non-empty finite sets have a choice function) we have $\mathfrak p \cdot \mathfrak n = \mathfrak p \cdot \mathfrak m \implies \mathfrak n = \mathfrak m$ for all non-zero Dedkind finite $\mathfrak{p,n,m}$, even without linear order.
Gershon Sageev claims that Ellentuck proved the above without the use ACF, but I couldn't find such proof
