Theorem cbvralsvw

Description: Change bound variable by using a substitution. Version of cbvralsv with a disjoint variable condition, which does not require ax-13 . (Contributed by NM, 20-Nov-2005) Avoid ax-13 . (Revised by GG, 10-Jan-2024) (Proof shortened by Wolf Lammen, 8-Mar-2025) Avoid ax-10 , ax-12 . (Revised by SN, 21-Aug-2025)

	Ref	Expression
Assertion	cbvralsvw	$$⊢\forall x\in A\phi \leftrightarrow \forall y\in A\left[y/x\right]\phi$$

Proof

Step	Hyp	Ref	Expression
1		sb8v	$$⊢\forall x\left(x\in A\to \phi \right)\leftrightarrow \forall y\left[y/x\right]\left(x\in A\to \phi \right)$$
2		df-ral	$$⊢\forall x\in A\phi \leftrightarrow \forall x\left(x\in A\to \phi \right)$$
3		df-ral	$$⊢\forall y\in A\left[y/x\right]\phi \leftrightarrow \forall y\left(y\in A\to \left[y/x\right]\phi \right)$$
4		eleq1w	$$⊢x=y\to \left(x\in A\leftrightarrow y\in A\right)$$
5	4	imbi1d	$$⊢x=y\to \left(\left(x\in A\to \phi \right)\leftrightarrow \left(y\in A\to \phi \right)\right)$$
6	5	sbbiiev	$$⊢\left[y/x\right]\left(x\in A\to \phi \right)\leftrightarrow \left[y/x\right]\left(y\in A\to \phi \right)$$
7		sbrimvw	$$⊢\left[y/x\right]\left(y\in A\to \phi \right)\leftrightarrow \left(y\in A\to \left[y/x\right]\phi \right)$$
8	6 7	bitr2i	$$⊢\left(y\in A\to \left[y/x\right]\phi \right)\leftrightarrow \left[y/x\right]\left(x\in A\to \phi \right)$$
9	8	albii	$$⊢\forall y\left(y\in A\to \left[y/x\right]\phi \right)\leftrightarrow \forall y\left[y/x\right]\left(x\in A\to \phi \right)$$
10	3 9	bitri	$$⊢\forall y\in A\left[y/x\right]\phi \leftrightarrow \forall y\left[y/x\right]\left(x\in A\to \phi \right)$$
11	1 2 10	3bitr4i	$$⊢\forall x\in A\phi \leftrightarrow \forall y\in A\left[y/x\right]\phi$$