Theorem sb3an

Description: Threefold conjunction inside and outside of a substitution are equivalent. (Contributed by NM, 14-Dec-2006)

	Ref	Expression
Assertion	sb3an	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) ↔ ( [ 𝑦 / 𝑥 ] 𝜑 ∧ [ 𝑦 / 𝑥 ] 𝜓 ∧ [ 𝑦 / 𝑥 ] 𝜒 ) )

Proof

Step	Hyp	Ref	Expression
1		df-3an	⊢ ( ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) ↔ ( ( 𝜑 ∧ 𝜓 ) ∧ 𝜒 ) )
2	1	sbbii	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) ↔ [ 𝑦 / 𝑥 ] ( ( 𝜑 ∧ 𝜓 ) ∧ 𝜒 ) )
3		sban	⊢ ( [ 𝑦 / 𝑥 ] ( ( 𝜑 ∧ 𝜓 ) ∧ 𝜒 ) ↔ ( [ 𝑦 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ∧ [ 𝑦 / 𝑥 ] 𝜒 ) )
4		sban	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ↔ ( [ 𝑦 / 𝑥 ] 𝜑 ∧ [ 𝑦 / 𝑥 ] 𝜓 ) )
5	4	anbi1i	⊢ ( ( [ 𝑦 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ∧ [ 𝑦 / 𝑥 ] 𝜒 ) ↔ ( ( [ 𝑦 / 𝑥 ] 𝜑 ∧ [ 𝑦 / 𝑥 ] 𝜓 ) ∧ [ 𝑦 / 𝑥 ] 𝜒 ) )
6		df-3an	⊢ ( ( [ 𝑦 / 𝑥 ] 𝜑 ∧ [ 𝑦 / 𝑥 ] 𝜓 ∧ [ 𝑦 / 𝑥 ] 𝜒 ) ↔ ( ( [ 𝑦 / 𝑥 ] 𝜑 ∧ [ 𝑦 / 𝑥 ] 𝜓 ) ∧ [ 𝑦 / 𝑥 ] 𝜒 ) )
7	5 6	bitr4i	⊢ ( ( [ 𝑦 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ∧ [ 𝑦 / 𝑥 ] 𝜒 ) ↔ ( [ 𝑦 / 𝑥 ] 𝜑 ∧ [ 𝑦 / 𝑥 ] 𝜓 ∧ [ 𝑦 / 𝑥 ] 𝜒 ) )
8	2 3 7	3bitri	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) ↔ ( [ 𝑦 / 𝑥 ] 𝜑 ∧ [ 𝑦 / 𝑥 ] 𝜓 ∧ [ 𝑦 / 𝑥 ] 𝜒 ) )