Theorem sbeqi

Description: Equality deduction for substitution. (Contributed by Giovanni Mascellani, 10-Apr-2018)

	Ref	Expression
Assertion	sbeqi	⊢ ( ( 𝑥 = 𝑦 ∧ ∀ 𝑧 ( 𝜑 ↔ 𝜓 ) ) → ( [ 𝑥 / 𝑧 ] 𝜑 ↔ [ 𝑦 / 𝑧 ] 𝜓 ) )

Proof

Step	Hyp	Ref	Expression
1		spsbbi	⊢ ( ∀ 𝑧 ( 𝜑 ↔ 𝜓 ) → ( [ 𝑥 / 𝑧 ] 𝜑 ↔ [ 𝑥 / 𝑧 ] 𝜓 ) )
2		sbequ	⊢ ( 𝑥 = 𝑦 → ( [ 𝑥 / 𝑧 ] 𝜓 ↔ [ 𝑦 / 𝑧 ] 𝜓 ) )
3	1 2	sylan9bbr	⊢ ( ( 𝑥 = 𝑦 ∧ ∀ 𝑧 ( 𝜑 ↔ 𝜓 ) ) → ( [ 𝑥 / 𝑧 ] 𝜑 ↔ [ 𝑦 / 𝑧 ] 𝜓 ) )