Theorem ralseu1d

Description: Deduction rule: Given "all some one" applied to a class, you can extract the "for all" part. (Contributed by David A. Wheeler, 21-Jul-2026)

	Ref	Expression
Hypothesis	ralseu1d.1	$$⊢\phi \to \forall \exists !x\in A\left(\psi \to \chi \right)$$
Assertion	ralseu1d	$$⊢\phi \to \forall x\in A\left(\psi \to \chi \right)$$

Proof

Step	Hyp	Ref	Expression
1		ralseu1d.1	$$⊢\phi \to \forall \exists !x\in A\left(\psi \to \chi \right)$$
2		df-ralseu	$$⊢\forall \exists !x\in A\left(\psi \to \chi \right)\leftrightarrow \left(\forall x\in A\left(\psi \to \chi \right)\wedge \exists !x\in A\psi \right)$$
3	1 2	sylib	$$⊢\phi \to \left(\forall x\in A\left(\psi \to \chi \right)\wedge \exists !x\in A\psi \right)$$
4	3	simpld	$$⊢\phi \to \forall x\in A\left(\psi \to \chi \right)$$