Theorem eliminable-abeqv

Description: A theorem used to prove the base case of the Eliminability Theorem (see section comment): abstraction equals variable. (Contributed by BJ, 30-Apr-2024) Beware not to use symmetry of class equality. (Proof modification is discouraged.) (New usage is discouraged.)

	Ref	Expression
Assertion	eliminable-abeqv	⊢ ( { 𝑥 ∣ 𝜑 } = 𝑦 ↔ ∀ 𝑧 ( [ 𝑧 / 𝑥 ] 𝜑 ↔ 𝑧 ∈ 𝑦 ) )

Proof

Step	Hyp	Ref	Expression
1		dfcleq	⊢ ( { 𝑥 ∣ 𝜑 } = 𝑦 ↔ ∀ 𝑧 ( 𝑧 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝑧 ∈ 𝑦 ) )
2		eliminable-velab	⊢ ( 𝑧 ∈ { 𝑥 ∣ 𝜑 } ↔ [ 𝑧 / 𝑥 ] 𝜑 )
3	2	bibi1i	⊢ ( ( 𝑧 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝑧 ∈ 𝑦 ) ↔ ( [ 𝑧 / 𝑥 ] 𝜑 ↔ 𝑧 ∈ 𝑦 ) )
4	3	albii	⊢ ( ∀ 𝑧 ( 𝑧 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝑧 ∈ 𝑦 ) ↔ ∀ 𝑧 ( [ 𝑧 / 𝑥 ] 𝜑 ↔ 𝑧 ∈ 𝑦 ) )
5	1 4	bitri	⊢ ( { 𝑥 ∣ 𝜑 } = 𝑦 ↔ ∀ 𝑧 ( [ 𝑧 / 𝑥 ] 𝜑 ↔ 𝑧 ∈ 𝑦 ) )