Theorem sban

Description: Conjunction inside and outside of a substitution are equivalent. Compare 19.26 . (Contributed by NM, 14-May-1993) (Proof shortened by Steven Nguyen, 13-Aug-2023)

	Ref	Expression
Assertion	sban	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ↔ ( [ 𝑦 / 𝑥 ] 𝜑 ∧ [ 𝑦 / 𝑥 ] 𝜓 ) )

Proof

Step	Hyp	Ref	Expression
1		simpl	⊢ ( ( 𝜑 ∧ 𝜓 ) → 𝜑 )
2	1	sbimi	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) → [ 𝑦 / 𝑥 ] 𝜑 )
3		simpr	⊢ ( ( 𝜑 ∧ 𝜓 ) → 𝜓 )
4	3	sbimi	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) → [ 𝑦 / 𝑥 ] 𝜓 )
5	2 4	jca	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) → ( [ 𝑦 / 𝑥 ] 𝜑 ∧ [ 𝑦 / 𝑥 ] 𝜓 ) )
6		pm3.2	⊢ ( 𝜑 → ( 𝜓 → ( 𝜑 ∧ 𝜓 ) ) )
7	6	sb2imi	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 → ( [ 𝑦 / 𝑥 ] 𝜓 → [ 𝑦 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ) )
8	7	imp	⊢ ( ( [ 𝑦 / 𝑥 ] 𝜑 ∧ [ 𝑦 / 𝑥 ] 𝜓 ) → [ 𝑦 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) )
9	5 8	impbii	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ↔ ( [ 𝑦 / 𝑥 ] 𝜑 ∧ [ 𝑦 / 𝑥 ] 𝜓 ) )