Theorem alsconv

Description: There is an equivalence between the two "all some" forms. (Contributed by David A. Wheeler, 22-Oct-2018)

	Ref	Expression
Assertion	alsconv	$$⊢\forall !x\left(x\in A\to \phi \right)\leftrightarrow \forall !x\in A\phi$$

Proof

Step	Hyp	Ref	Expression
1		df-ral	$$⊢\forall x\in A\phi \leftrightarrow \forall x\left(x\in A\to \phi \right)$$
2	1	anbi1i	$$⊢\left(\forall x\in A\phi \wedge \exists xx\in A\right)\leftrightarrow \left(\forall x\left(x\in A\to \phi \right)\wedge \exists xx\in A\right)$$
3		df-alsc	$$⊢\forall !x\in A\phi \leftrightarrow \left(\forall x\in A\phi \wedge \exists xx\in A\right)$$
4		df-alsi	$$⊢\forall !x\left(x\in A\to \phi \right)\leftrightarrow \left(\forall x\left(x\in A\to \phi \right)\wedge \exists xx\in A\right)$$
5	2 3 4	3bitr4ri	$$⊢\forall !x\left(x\in A\to \phi \right)\leftrightarrow \forall !x\in A\phi$$