Theorem eliminable-veqab

Description: A theorem used to prove the base case of the Eliminability Theorem (see section comment): variable equals abstraction. (Contributed by BJ, 30-Apr-2024) (Proof modification is discouraged.) (New usage is discouraged.)

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

Proof

Step	Hyp	Ref	Expression
1		dfcleq	⊢ ( 𝑥 = { 𝑦 ∣ 𝜑 } ↔ ∀ 𝑧 ( 𝑧 ∈ 𝑥 ↔ 𝑧 ∈ { 𝑦 ∣ 𝜑 } ) )
2		eliminable-velab	⊢ ( 𝑧 ∈ { 𝑦 ∣ 𝜑 } ↔ [ 𝑧 / 𝑦 ] 𝜑 )
3	2	bibi2i	⊢ ( ( 𝑧 ∈ 𝑥 ↔ 𝑧 ∈ { 𝑦 ∣ 𝜑 } ) ↔ ( 𝑧 ∈ 𝑥 ↔ [ 𝑧 / 𝑦 ] 𝜑 ) )
4	3	albii	⊢ ( ∀ 𝑧 ( 𝑧 ∈ 𝑥 ↔ 𝑧 ∈ { 𝑦 ∣ 𝜑 } ) ↔ ∀ 𝑧 ( 𝑧 ∈ 𝑥 ↔ [ 𝑧 / 𝑦 ] 𝜑 ) )
5	1 4	bitri	⊢ ( 𝑥 = { 𝑦 ∣ 𝜑 } ↔ ∀ 𝑧 ( 𝑧 ∈ 𝑥 ↔ [ 𝑧 / 𝑦 ] 𝜑 ) )