Theorem sbcan

Description: Distribution of class substitution over conjunction. (Contributed by NM, 31-Dec-2016) (Revised by NM, 17-Aug-2018)

	Ref	Expression
Assertion	sbcan	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\left(\phi \wedge \psi \right)\leftrightarrow \left(\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \wedge \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\psi \right)$$

Proof

Step	Hyp	Ref	Expression
1		sbcex	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\left(\phi \wedge \psi \right)\to A\in \mathrm{V}$$
2		sbcex	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\psi \to A\in \mathrm{V}$$
3	2	adantl	$$⊢\left(\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \wedge \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\psi \right)\to A\in \mathrm{V}$$
4		dfsbcq2	$$⊢y=A\to \left(\left[y/x\right]\left(\phi \wedge \psi \right)\leftrightarrow \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\left(\phi \wedge \psi \right)\right)$$
5		dfsbcq2	$$⊢y=A\to \left(\left[y/x\right]\phi \leftrightarrow \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \right)$$
6		dfsbcq2	$$⊢y=A\to \left(\left[y/x\right]\psi \leftrightarrow \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\psi \right)$$
7	5 6	anbi12d	$$⊢y=A\to \left(\left(\left[y/x\right]\phi \wedge \left[y/x\right]\psi \right)\leftrightarrow \left(\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \wedge \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\psi \right)\right)$$
8		sban	$$⊢\left[y/x\right]\left(\phi \wedge \psi \right)\leftrightarrow \left(\left[y/x\right]\phi \wedge \left[y/x\right]\psi \right)$$
9	4 7 8	vtoclbg	$$⊢A\in \mathrm{V}\to \left(\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\left(\phi \wedge \psi \right)\leftrightarrow \left(\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \wedge \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\psi \right)\right)$$
10	1 3 9	pm5.21nii	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\left(\phi \wedge \psi \right)\leftrightarrow \left(\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \wedge \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\psi \right)$$