Bounded finite set theory is:
Extensionality + $\Delta_0$-Separation + Adjunction?
Where Adjunction is: $\forall x \forall y \ (x \cup \{y\} \text { exists})$
Would this theory be equivalent to bounded arithmetic?
Bounded finite set theory is:
Extensionality + $\Delta_0$-Separation + Adjunction?
Where Adjunction is: $\forall x \forall y \ (x \cup \{y\} \text { exists})$
Would this theory be equivalent to bounded arithmetic?
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