Theorem eliminable-abeqab 36891

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.)

Assertion

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

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

Allowed substitution hints:   𝜑(𝑥,𝑦)   𝜓(𝑥,𝑦)

Proof of Theorem eliminable-abeqab

Step	Hyp	Ref	Expression
1		dfcleq 2729	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑦 ∣ 𝜓} ↔ ∀𝑧(𝑧 ∈ {𝑥 ∣ 𝜑} ↔ 𝑧 ∈ {𝑦 ∣ 𝜓}))
2		eliminable-velab 36888	. . . 4 ⊢ (𝑧 ∈ {𝑥 ∣ 𝜑} ↔ [𝑧 / 𝑥]𝜑)
3		eliminable-velab 36888	. . . 4 ⊢ (𝑧 ∈ {𝑦 ∣ 𝜓} ↔ [𝑧 / 𝑦]𝜓)
4	2, 3	bibi12i 339	. . 3 ⊢ ((𝑧 ∈ {𝑥 ∣ 𝜑} ↔ 𝑧 ∈ {𝑦 ∣ 𝜓}) ↔ ([𝑧 / 𝑥]𝜑 ↔ [𝑧 / 𝑦]𝜓))
5	4	albii 1819	. 2 ⊢ (∀𝑧(𝑧 ∈ {𝑥 ∣ 𝜑} ↔ 𝑧 ∈ {𝑦 ∣ 𝜓}) ↔ ∀𝑧([𝑧 / 𝑥]𝜑 ↔ [𝑧 / 𝑦]𝜓))
6	1, 5	bitri 275	1 ⊢ ({𝑥 ∣ 𝜑} = {𝑦 ∣ 𝜓} ↔ ∀𝑧([𝑧 / 𝑥]𝜑 ↔ [𝑧 / 𝑦]𝜓))

Syntax hints:   ↔ wb 206  ∀wal 1538   = wceq 1540  [wsb 2065   ∈ wcel 2109  {cab 2714

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-9 2119  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1780  df-clab 2715  df-cleq 2728

This theorem is referenced by: (None)