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	⊢ ( [ 𝐴 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ↔ ( [ 𝐴 / 𝑥 ] 𝜑 ∧ [ 𝐴 / 𝑥 ] 𝜓 ) )

Proof

Step	Hyp	Ref	Expression
1		sbcex	⊢ ( [ 𝐴 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) → 𝐴 ∈ V )
2		sbcex	⊢ ( [ 𝐴 / 𝑥 ] 𝜓 → 𝐴 ∈ V )
3	2	adantl	⊢ ( ( [ 𝐴 / 𝑥 ] 𝜑 ∧ [ 𝐴 / 𝑥 ] 𝜓 ) → 𝐴 ∈ V )
4		dfsbcq2	⊢ ( 𝑦 = 𝐴 → ( [ 𝑦 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ↔ [ 𝐴 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ) )
5		dfsbcq2	⊢ ( 𝑦 = 𝐴 → ( [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝐴 / 𝑥 ] 𝜑 ) )
6		dfsbcq2	⊢ ( 𝑦 = 𝐴 → ( [ 𝑦 / 𝑥 ] 𝜓 ↔ [ 𝐴 / 𝑥 ] 𝜓 ) )
7	5 6	anbi12d	⊢ ( 𝑦 = 𝐴 → ( ( [ 𝑦 / 𝑥 ] 𝜑 ∧ [ 𝑦 / 𝑥 ] 𝜓 ) ↔ ( [ 𝐴 / 𝑥 ] 𝜑 ∧ [ 𝐴 / 𝑥 ] 𝜓 ) ) )
8		sban	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ↔ ( [ 𝑦 / 𝑥 ] 𝜑 ∧ [ 𝑦 / 𝑥 ] 𝜓 ) )
9	4 7 8	vtoclbg	⊢ ( 𝐴 ∈ V → ( [ 𝐴 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ↔ ( [ 𝐴 / 𝑥 ] 𝜑 ∧ [ 𝐴 / 𝑥 ] 𝜓 ) ) )
10	1 3 9	pm5.21nii	⊢ ( [ 𝐴 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ↔ ( [ 𝐴 / 𝑥 ] 𝜑 ∧ [ 𝐴 / 𝑥 ] 𝜓 ) )