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Every nonempty finite set $x$ has an $\in$-maximal element, a set $z\in x$ with $z\notin u$ for any $u\in x$, since one can simply climb up via $\in$ inside $x$ until one reaches a maximal element.

But the statement is not true of all infinite sets, since the set HF of hereditarily finite sets itself, for example, has no $\in$-maximal elements.

Using this idea, we can produce a counterexample to your transfer principles. Let $\phi(y,x)$ assert that $x$ is not empty and if $x$ has an $\in$-maximal element, then $y\in x$. This can be expressed by a comparatively short formula. Namely, let $\phi(y,x)$ be the assertion $$(\exists z\ z\in x)\wedge[(\exists z\in x\ \forall u\in x\ z\notin u)\to y\in x].$$

If $x\in\text{HF}$ and $\phi(y,x)$, then $x$ is not empty and $y\in x$, since every nonempty element of $\text{HF}$ has $\in$-maximal elements. But there are sets $x$ in $V$ which are not empty, but have no $\in$-maximal elements, such as $x=\omega+\omega$, and in this case $\phi(y,x)$ holds for all $y$, including the case where $y$ is a proper class.

So this violates your transfer principles.

Your transfer principle contradicts the axiom of foundation.

To see this, observe that under foundation, every nonempty finite set $x$ has an $\in$-maximal element, a set $z\in x$ with $z\notin u$ for any $u\in x$, since one can simply climb up via $\in$ inside $x$ until one reaches a maximal element.

But not every infinite set is like that, since the set HF of hereditarily finite sets itself, for example, has no $\in$-maximal elements.

Using this idea, we can produce a counterexample to your transfer principles under foundation. Let $\phi(y,x)$ assert that $x$ is not empty and if $x$ has an $\in$-maximal element, then $y\in x$. This can be expressed by a comparatively short formula. Namely, let $\phi(y,x)$ be the assertion $$(\exists z\ z\in x)\wedge[(\exists z\in x\ \forall u\in x\ z\notin u)\to y\in x].$$

If $x\in\text{HF}$ and $\phi(y,x)$, then $x$ is not empty and $y\in x$, since every nonempty element of $\text{HF}$ has $\in$-maximal elements. But there are sets $x$ in $V$ which are not empty, but have no $\in$-maximal elements, such as $x=\omega+\omega$, and in this case $\phi(y,x)$ holds for all $y$, including the case where $y$ is a proper class.

So this violates your transfer principles.

Every nonempty finite set $x$ has an $\in$-maximal element $y\in x$, since one can simply climb up via $\in$ inside $x$ until one reaches a maximal element.

But the statement is not true of all infinite sets, since the set HF of hereditarily finite sets itself, for example, has no  $\in$-maximal elements.

Therefore, the property $\varphi(x)$ asserting: $$x\neq\emptyset\to \exists y\in x\ \forall z\in x\ (y\notin z)$$ is true of every finite set, but it does not transfer to the infinite realm.

Since this assertion is very short, it would seem to be a violation of your transfer principles. But I'm not actually sure, since I haven't followed your axioms and notation in detail.

Every nonempty finite set $x$ has an $\in$-maximal element, a set $z\in x$ with $z\notin u$ for any $u\in x$, since one can simply climb up via $\in$ inside $x$ until one reaches a maximal element.

But the statement is not true of all infinite sets, since the set HF of hereditarily finite sets itself, for example, has no  $\in$-maximal elements.

Using this idea, we can produce a counterexample to your transfer principles. Let $\phi(y,x)$ assert that $x$ is not empty and if $x$ has an $\in$-maximal element, then $y\in x$. This can be expressed by a comparatively short formula. Namely, let $\phi(y,x)$ be the assertion $$(\exists z\ z\in x)\wedge[(\exists z\in x\ \forall u\in x\ z\notin u)\to y\in x].$$

If $x\in\text{HF}$ and $\phi(y,x)$, then $x$ is not empty and $y\in x$, since every nonempty element of $\text{HF}$ has $\in$-maximal elements. But there are sets $x$ in $V$ which are not empty, but have no $\in$-maximal elements, such as $x=\omega+\omega$, and in this case $\phi(y,x)$ holds for all $y$, including the case where $y$ is a proper class.

So this violates your transfer principles.

Every nonempty finite set $x$ has an $\in$-maximal element $y\in x$, since one can simply climb up via $\in$ inside $x$ until one reaches a maximal element.

But the statement is not true of all infinite sets, since the usual ordinal $\omega$, for example, is closed under the singleton operation and therefore has no $\in$-maximal elements.

Therefore, the property $\varphi(x)$ asserting: $$x\neq\emptyset\to \exists y\in x\ \forall z\in x\ (y\notin z)$$ is true of every finite set, but it does not transfer to the infinite realm.

Since this assertion is very short, it would seem to be a violation of your transfer principles. But I'm not actually sure, since I haven't followed your axioms and notation in detail.

Every nonempty finite set $x$ has an $\in$-maximal element $y\in x$, since one can simply climb up via $\in$ inside $x$ until one reaches a maximal element.

But the statement is not true of all infinite sets, since the set HF of hereditarily finite sets itself, for example, has no $\in$-maximal elements.

Therefore, the property $\varphi(x)$ asserting: $$x\neq\emptyset\to \exists y\in x\ \forall z\in x\ (y\notin z)$$ is true of every finite set, but it does not transfer to the infinite realm.

Since this assertion is very short, it would seem to be a violation of your transfer principles. But I'm not actually sure, since I haven't followed your axioms and notation in detail.