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Theorem cdeqab 3458
 Description: Distribute conditional equality over abstraction. (Contributed by Mario Carneiro, 11-Aug-2016.)
Hypothesis
Ref Expression
cdeqnot.1 CondEq(𝑥 = 𝑦 → (𝜑𝜓))
Assertion
Ref Expression
cdeqab CondEq(𝑥 = 𝑦 → {𝑧𝜑} = {𝑧𝜓})
Distinct variable groups:   𝑥,𝑧   𝑦,𝑧
Allowed substitution hints:   𝜑(𝑥,𝑦,𝑧)   𝜓(𝑥,𝑦,𝑧)

Proof of Theorem cdeqab
StepHypRef Expression
1 cdeqnot.1 . . . 4 CondEq(𝑥 = 𝑦 → (𝜑𝜓))
21cdeqri 3454 . . 3 (𝑥 = 𝑦 → (𝜑𝜓))
32abbidv 2770 . 2 (𝑥 = 𝑦 → {𝑧𝜑} = {𝑧𝜓})
43cdeqi 3453 1 CondEq(𝑥 = 𝑦 → {𝑧𝜑} = {𝑧𝜓})
 Colors of variables: wff setvar class Syntax hints:   ↔ wb 196   = wceq 1523  {cab 2637  CondEqwcdeq 3451 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1762  ax-4 1777  ax-5 1879  ax-6 1945  ax-7 1981  ax-9 2039  ax-10 2059  ax-11 2074  ax-12 2087  ax-13 2282  ax-ext 2631 This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-tru 1526  df-ex 1745  df-nf 1750  df-sb 1938  df-clab 2638  df-cleq 2644  df-clel 2647  df-cdeq 3452 This theorem is referenced by: (None)
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