Theorem sblbis

Description: Introduce left biconditional inside of a substitution. (Contributed by NM, 19-Aug-1993)

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

Proof

Step	Hyp	Ref	Expression
1		sblbis.1	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ 𝜓 )
2		sbbi	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜒 ↔ 𝜑 ) ↔ ( [ 𝑦 / 𝑥 ] 𝜒 ↔ [ 𝑦 / 𝑥 ] 𝜑 ) )
3	1	bibi2i	⊢ ( ( [ 𝑦 / 𝑥 ] 𝜒 ↔ [ 𝑦 / 𝑥 ] 𝜑 ) ↔ ( [ 𝑦 / 𝑥 ] 𝜒 ↔ 𝜓 ) )
4	2 3	bitri	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜒 ↔ 𝜑 ) ↔ ( [ 𝑦 / 𝑥 ] 𝜒 ↔ 𝜓 ) )