Equivalence between a set and the subset of its subset

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The discussion centers on the equivalence relation between a set A and its subset A1, specifically questioning whether A's lack of equivalence to A1 implies A's lack of equivalence to any subset of A1. It is established that the truth of this statement depends on the chosen equivalence relation, such as bijection or other criteria like parity. When using bijections, the statement holds true; however, under different equivalence relations, it may not. The conversation reflects on the complexities of understanding these relationships, referencing Cantor-Bernstein's theorem as a learning experience. Ultimately, the nuances of equivalence relations significantly influence the validity of the original claim.
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Is it true that: If A is not equivalent to its subset A1. Then A is not equivalent to any subset of A1?
 
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You should consider also informing us about the relevant equivalence relation, as the truth or falsity of your statement heavily depends on that information.

For example, if you for your eq.rel. use existence of a bijection, the statement is true, but other relations, like having same parity, will render your statement false.
 
actually this is trivial. I just learned all this the complicated way, .i.e. proving cantor bernstein without the well ordering theorem.
 
I'm taking a look at intuitionistic propositional logic (IPL). Basically it exclude Double Negation Elimination (DNE) from the set of axiom schemas replacing it with Ex falso quodlibet: ⊥ → p for any proposition p (including both atomic and composite propositions). In IPL, for instance, the Law of Excluded Middle (LEM) p ∨ ¬p is no longer a theorem. My question: aside from the logic formal perspective, is IPL supposed to model/address some specific "kind of world" ? Thanks.
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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