Theorem eliminable-veqab 34586

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.)

Assertion

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

Distinct variable groups:   𝑥,𝑧   𝑦,𝑧   𝜑,𝑧

Allowed substitution hints:   𝜑(𝑥,𝑦)

Proof of Theorem eliminable-veqab

Step	Hyp	Ref	Expression
1		dfcleq 2752	. 2 ⊢ (𝑥 = {𝑦 ∣ 𝜑} ↔ ∀𝑧(𝑧 ∈ 𝑥 ↔ 𝑧 ∈ {𝑦 ∣ 𝜑}))
2		eliminable-velab 34585	. . . 4 ⊢ (𝑧 ∈ {𝑦 ∣ 𝜑} ↔ [𝑧 / 𝑦]𝜑)
3	2	bibi2i 342	. . 3 ⊢ ((𝑧 ∈ 𝑥 ↔ 𝑧 ∈ {𝑦 ∣ 𝜑}) ↔ (𝑧 ∈ 𝑥 ↔ [𝑧 / 𝑦]𝜑))
4	3	albii 1822	. 2 ⊢ (∀𝑧(𝑧 ∈ 𝑥 ↔ 𝑧 ∈ {𝑦 ∣ 𝜑}) ↔ ∀𝑧(𝑧 ∈ 𝑥 ↔ [𝑧 / 𝑦]𝜑))
5	1, 4	bitri 278	1 ⊢ (𝑥 = {𝑦 ∣ 𝜑} ↔ ∀𝑧(𝑧 ∈ 𝑥 ↔ [𝑧 / 𝑦]𝜑))

Syntax hints:   ↔ wb 209  ∀wal 1537   = wceq 1539  [wsb 2070   ∈ wcel 2112  {cab 2736

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1912  ax-6 1971  ax-7 2016  ax-9 2122  ax-ext 2730

This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1783  df-clab 2737  df-cleq 2751

This theorem is referenced by:  eliminable-abelv  34589  eliminable-abelab  34590