Theorem sbcalf

Description: Move universal quantifier in and out of class substitution, with an explicit non-free variable condition. (Contributed by Giovanni Mascellani, 29-May-2019)

	Ref	Expression
Hypothesis	sbcalf.1	$$⊢\underset{\_}{Ⅎ}yA$$
Assertion	sbcalf	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\forall y\phi \leftrightarrow \forall y\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi$$

Proof

Step	Hyp	Ref	Expression
1		sbcalf.1	$$⊢\underset{\_}{Ⅎ}yA$$
2		nfv	$$⊢Ⅎz\phi$$
3	2	sb8v	$$⊢\forall y\phi \leftrightarrow \forall z\left[z/y\right]\phi$$
4	3	sbcbii	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\forall y\phi \leftrightarrow \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\forall z\left[z/y\right]\phi$$
5		sbcal	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\forall z\left[z/y\right]\phi \leftrightarrow \forall z\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\left[z/y\right]\phi$$
6		nfs1v	$$⊢Ⅎy\left[z/y\right]\phi$$
7	1 6	nfsbcw	$$⊢Ⅎy\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\left[z/y\right]\phi$$
8		nfv	$$⊢Ⅎz\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi$$
9		sbequ12r	$$⊢z=y\to \left(\left[z/y\right]\phi \leftrightarrow \phi \right)$$
10	9	sbcbidv	$$⊢z=y\to \left(\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\left[z/y\right]\phi \leftrightarrow \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \right)$$
11	7 8 10	cbvalv1	$$⊢\forall z\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\left[z/y\right]\phi \leftrightarrow \forall y\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi$$
12	4 5 11	3bitri	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\forall y\phi \leftrightarrow \forall y\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi$$