Theorem eliminable-abeqab

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

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

Proof

Step	Hyp	Ref	Expression
1		dfcleq	⊢ ( { 𝑥 ∣ 𝜑 } = { 𝑦 ∣ 𝜓 } ↔ ∀ 𝑧 ( 𝑧 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝑧 ∈ { 𝑦 ∣ 𝜓 } ) )
2		eliminable-velab	⊢ ( 𝑧 ∈ { 𝑥 ∣ 𝜑 } ↔ [ 𝑧 / 𝑥 ] 𝜑 )
3		eliminable-velab	⊢ ( 𝑧 ∈ { 𝑦 ∣ 𝜓 } ↔ [ 𝑧 / 𝑦 ] 𝜓 )
4	2 3	bibi12i	⊢ ( ( 𝑧 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝑧 ∈ { 𝑦 ∣ 𝜓 } ) ↔ ( [ 𝑧 / 𝑥 ] 𝜑 ↔ [ 𝑧 / 𝑦 ] 𝜓 ) )
5	4	albii	⊢ ( ∀ 𝑧 ( 𝑧 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝑧 ∈ { 𝑦 ∣ 𝜓 } ) ↔ ∀ 𝑧 ( [ 𝑧 / 𝑥 ] 𝜑 ↔ [ 𝑧 / 𝑦 ] 𝜓 ) )
6	1 5	bitri	⊢ ( { 𝑥 ∣ 𝜑 } = { 𝑦 ∣ 𝜓 } ↔ ∀ 𝑧 ( [ 𝑧 / 𝑥 ] 𝜑 ↔ [ 𝑧 / 𝑦 ] 𝜓 ) )