If $\varphi, \psi$ are two parameter free formulas in the language of set theory $T$ such that there is a theorem of $T$ that decides on the exact cardinal comparison of the sets they define, i.e. we have: $ (T \vdash|S^{\varphi}|>|S^\psi|)$ or $( T \vdash|S^{\varphi}|=|S^\psi|)$ or $( T \vdash|S^{\varphi}|<|S^\psi|)$, (where $S^{\phi}$ is the set definable after formula $\phi$), then we'd say that formulas $\varphi, \psi$ invoke mutually decidable cardinal comparisons in $T$, I'll denote that by $``|S^{\varphi}|,|S^{\psi}| \text { decided in } T"$

Lets also define a notion of cardinal subdecidablility between theories from the same language, of course with respect to cardinal comparison's in them, so Lets say that $$T \text{ cardinally subdecidable to } H \iff \\ \forall \varphi, \psi (|S^{\varphi}|,|S^{\psi}|\text{ decided in } T \to |S^{\varphi}|,|S^{\psi}|\text{ decided in } H) $$

Now we have $\text{ZFC}$ being cardinally subdecidable to $\text{ ZF+ V=L}$, but not the converse, so as far as cardinal decidability is concerned the latter is stronger than the former.

Question: Can there be a theory $T$ in the language of $\text{ZF}$ that is equi-consistent with $\text{ ZF+ V=L}$ and yet has $\text{ ZF+ V=L} \text{ cardinally subdecidable to } T$ but not the converse ?

{EDIT: this posting has been edited, the additional text is in italics}

If $\varphi, \psi$ are two parameter free formulas in the language of set theory $T$ such that there is a theorem of $T$ that decides on the exact cardinal comparison of the sets they define, i.e. we have: $ (T \vdash|S^{\varphi}|>|S^\psi|)$ or $( T \vdash|S^{\varphi}|=|S^\psi|)$ or $( T \vdash|S^{\varphi}|<|S^\psi|)$, (where $S^{\phi}$ is the set definable after formula $\phi$), then we'd say that formulas $\varphi, \psi$ invoke mutually decidable cardinal comparisons in $T$, I'll denote that by $``|S^{\varphi}|,|S^{\psi}| \text { decided in } T"$

Lets also define a notion of cardinal subdecidablility between theories from the same language, of course with respect to cardinal comparison's in them, so Lets say that $$T \text{ cardinally subdecidable to } H \iff \\ \forall \varphi, \psi (|S^{\varphi}|,|S^{\psi}|\text{ decided in } T \to |S^{\varphi}|,|S^{\psi}|\text{ decided in } H) $$

Now we have $\text{ZFC}$ being cardinally subdecidable to $\text{ ZF+ V=L}$, but not the converse, so as far as cardinal decidability is concerned the latter is stronger than the former. This is expected as $\text{ZFC}$ is already a subtheory of $\text{ ZF+ V=L}$, so this additional strength can be explained by this fact, but it doesn't seem to be merely so, since $\text{ ZF+ V=L}$ is more decidable on cardinal comparisons than its counterpart $\text{ ZF+ }V \neq L$, while if the additonal strength of cardinal decidability in $\text{ ZF+ V=L}$ over $\text{ZFC}$ was merely due to the latter being a subtheory of the former, one would have expected to have $\text{ ZF+ V=L} \text { cardinally equidecidable with } \text{ ZF+ } V \neq L$. From this observation we may define a new criterion for cardinal subdecidability of theories, that is: $$T \text{ truly cardinally subdecidable to } H \iff T \text{ cardinally subdecidable to } H \wedge \neg (H \text{ cardinally subdecidable to } T) \wedge H^{\neg} \text{ cardinally subdecidable to } H \wedge \neg(H \text{ cardinally subdecidable to } H^{\neg}) $$ Where $H^{\neg} = T + \neg (H-T)$, where $H-T$ are the axioms of $H$ that exceeds $T$.

Question: Can there be a theory $T$ in the language of $\text{ZF}$ that is equi-consistent with $\text{ ZF+ V=L}$ and yet has $\text{ ZF+ V=L} \text{ truly cardinally subdecidable to } T$?

If $\varphi, \psi$ are two parameter free formulas in the language of set theory $T$ such that there is a theorem of $T$ that decides on the exact cardinal comparison of the sets they define, i.e. we have: $ (T \vdash|S^{\varphi}|>|S^\psi|)$ or $( T \vdash|S^{\varphi}|=|S^\psi|)$ or $( T \vdash|S^{\varphi}|<|S^\psi|)$, (where $S^{\phi}$ is the set definable after formula $\phi$), then we'd say that formulas $\varphi, \psi$ invoke mutually decidable cardinal comparisons in $T$, I'll denote that by $``|S^{\varphi}|,|S^{\psi}| \text { decided in } T"$

Lets also define a notion of cardinal subdecidablility between theories from the same language, of course with respect to cardinal comparison's in them, so Lets say that $$T \text{ cardinally subdecidable to } H \iff \\ \forall \varphi, \psi (|S^{\varphi}|,|S^{\psi}|\text{ decided in } T \to |S^{\varphi}|,|S^{\psi}|\text{ decided in } H) $$

Now we have $\text{ZFC}$ being cardinally subdecidable to $\text{ ZF+ V=L}$ but not the converse, so as far as cardinal decidability is concerned the latter is stronger than the former.

Question: Can there be a theory $T$ that is equi-consistent with $\text{ ZF+ V=L}$ and yet has $\text{ ZF+ V=L} \text{ cardinally subdecidable to } T$?

If $\varphi, \psi$ are two parameter free formulas in the language of set theory $T$ such that there is a theorem of $T$ that decides on the exact cardinal comparison of the sets they define, i.e. we have: $ (T \vdash|S^{\varphi}|>|S^\psi|)$ or $( T \vdash|S^{\varphi}|=|S^\psi|)$ or $( T \vdash|S^{\varphi}|<|S^\psi|)$, (where $S^{\phi}$ is the set definable after formula $\phi$), then we'd say that formulas $\varphi, \psi$ invoke mutually decidable cardinal comparisons in $T$, I'll denote that by $``|S^{\varphi}|,|S^{\psi}| \text { decided in } T"$

Lets also define a notion of cardinal subdecidablility between theories from the same language, of course with respect to cardinal comparison's in them, so Lets say that $$T \text{ cardinally subdecidable to } H \iff \\ \forall \varphi, \psi (|S^{\varphi}|,|S^{\psi}|\text{ decided in } T \to |S^{\varphi}|,|S^{\psi}|\text{ decided in } H) $$

Now we have $\text{ZFC}$ being cardinally subdecidable to $\text{ ZF+ V=L}$, but not the converse, so as far as cardinal decidability is concerned the latter is stronger than the former.

Question: Can there be a theory $T$ in the language of $\text{ZF}$ that is equi-consistent with $\text{ ZF+ V=L}$ and yet has $\text{ ZF+ V=L} \text{ cardinally subdecidable to } T$ but not the converse ?