Theorem sb6a

Description: Equivalence for substitution. (Contributed by NM, 2-Jun-1993) (Proof shortened by Wolf Lammen, 23-Sep-2018)

	Ref	Expression
Assertion	sb6a	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ ∀ 𝑥 ( 𝑥 = 𝑦 → [ 𝑥 / 𝑦 ] 𝜑 ) )

Proof

Step	Hyp	Ref	Expression
1		sbcov	⊢ ( [ 𝑦 / 𝑥 ] [ 𝑥 / 𝑦 ] 𝜑 ↔ [ 𝑦 / 𝑥 ] 𝜑 )
2		sb6	⊢ ( [ 𝑦 / 𝑥 ] [ 𝑥 / 𝑦 ] 𝜑 ↔ ∀ 𝑥 ( 𝑥 = 𝑦 → [ 𝑥 / 𝑦 ] 𝜑 ) )
3	1 2	bitr3i	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ ∀ 𝑥 ( 𝑥 = 𝑦 → [ 𝑥 / 𝑦 ] 𝜑 ) )