Theorem sbrbis

Description: Introduce right biconditional inside of a substitution. (Contributed by NM, 18-Aug-1993)

	Ref	Expression
Hypothesis	sbrbis.1	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ 𝜓 )
Assertion	sbrbis	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 ↔ 𝜒 ) ↔ ( 𝜓 ↔ [ 𝑦 / 𝑥 ] 𝜒 ) )

Proof

Step	Hyp	Ref	Expression
1		sbrbis.1	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ 𝜓 )
2		sbbi	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 ↔ 𝜒 ) ↔ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝑦 / 𝑥 ] 𝜒 ) )
3	1	bibi1i	⊢ ( ( [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝑦 / 𝑥 ] 𝜒 ) ↔ ( 𝜓 ↔ [ 𝑦 / 𝑥 ] 𝜒 ) )
4	2 3	bitri	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 ↔ 𝜒 ) ↔ ( 𝜓 ↔ [ 𝑦 / 𝑥 ] 𝜒 ) )