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Theorem eliminable-abeqab 34572
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 ({𝑥𝜑} = {𝑦𝜓} ↔ ∀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
StepHypRef Expression
1 dfcleq 2752 . 2 ({𝑥𝜑} = {𝑦𝜓} ↔ ∀𝑧(𝑧 ∈ {𝑥𝜑} ↔ 𝑧 ∈ {𝑦𝜓}))
2 eliminable-velab 34569 . . . 4 (𝑧 ∈ {𝑥𝜑} ↔ [𝑧 / 𝑥]𝜑)
3 eliminable-velab 34569 . . . 4 (𝑧 ∈ {𝑦𝜓} ↔ [𝑧 / 𝑦]𝜓)
42, 3bibi12i 344 . . 3 ((𝑧 ∈ {𝑥𝜑} ↔ 𝑧 ∈ {𝑦𝜓}) ↔ ([𝑧 / 𝑥]𝜑 ↔ [𝑧 / 𝑦]𝜓))
54albii 1822 . 2 (∀𝑧(𝑧 ∈ {𝑥𝜑} ↔ 𝑧 ∈ {𝑦𝜓}) ↔ ∀𝑧([𝑧 / 𝑥]𝜑 ↔ [𝑧 / 𝑦]𝜓))
61, 5bitri 278 1 ({𝑥𝜑} = {𝑦𝜓} ↔ ∀𝑧([𝑧 / 𝑥]𝜑 ↔ [𝑧 / 𝑦]𝜓))
Colors of variables: wff setvar class
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: (None)
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