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Theorem eliminable-veqab 36235
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 = {𝑦𝜑} ↔ ∀𝑧(𝑧𝑥 ↔ [𝑧 / 𝑦]𝜑))
Distinct variable groups:   𝑥,𝑧   𝑦,𝑧   𝜑,𝑧
Allowed substitution hints:   𝜑(𝑥,𝑦)

Proof of Theorem eliminable-veqab
StepHypRef Expression
1 dfcleq 2717 . 2 (𝑥 = {𝑦𝜑} ↔ ∀𝑧(𝑧𝑥𝑧 ∈ {𝑦𝜑}))
2 eliminable-velab 36234 . . . 4 (𝑧 ∈ {𝑦𝜑} ↔ [𝑧 / 𝑦]𝜑)
32bibi2i 337 . . 3 ((𝑧𝑥𝑧 ∈ {𝑦𝜑}) ↔ (𝑧𝑥 ↔ [𝑧 / 𝑦]𝜑))
43albii 1813 . 2 (∀𝑧(𝑧𝑥𝑧 ∈ {𝑦𝜑}) ↔ ∀𝑧(𝑧𝑥 ↔ [𝑧 / 𝑦]𝜑))
51, 4bitri 275 1 (𝑥 = {𝑦𝜑} ↔ ∀𝑧(𝑧𝑥 ↔ [𝑧 / 𝑦]𝜑))
Colors of variables: wff setvar class
Syntax hints:  wb 205  wal 1531   = wceq 1533  [wsb 2059  wcel 2098  {cab 2701
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-9 2108  ax-ext 2695
This theorem depends on definitions:  df-bi 206  df-an 396  df-ex 1774  df-clab 2702  df-cleq 2716
This theorem is referenced by:  eliminable-abelv  36238  eliminable-abelab  36239
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