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Theorem cdeqab1 3766
Description: Distribute conditional equality over abstraction. Usage of this theorem is discouraged because it depends on ax-13 2372. (Contributed by Mario Carneiro, 11-Aug-2016.) (New usage is discouraged.)
Hypothesis
Ref Expression
cdeqnot.1 CondEq(𝑥 = 𝑦 → (𝜑𝜓))
Assertion
Ref Expression
cdeqab1 CondEq(𝑥 = 𝑦 → {𝑥𝜑} = {𝑦𝜓})
Distinct variable groups:   𝜓,𝑥   𝜑,𝑦
Allowed substitution hints:   𝜑(𝑥)   𝜓(𝑦)

Proof of Theorem cdeqab1
StepHypRef Expression
1 nfv 1918 . . 3 𝑦𝜑
2 nfv 1918 . . 3 𝑥𝜓
3 cdeqnot.1 . . . 4 CondEq(𝑥 = 𝑦 → (𝜑𝜓))
43cdeqri 3760 . . 3 (𝑥 = 𝑦 → (𝜑𝜓))
51, 2, 4cbvab 2809 . 2 {𝑥𝜑} = {𝑦𝜓}
65cdeqth 3761 1 CondEq(𝑥 = 𝑦 → {𝑥𝜑} = {𝑦𝜓})
Colors of variables: wff setvar class
Syntax hints:  wb 205   = wceq 1542  {cab 2710  CondEqwcdeq 3757
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 1914  ax-6 1972  ax-7 2012  ax-9 2117  ax-10 2138  ax-11 2155  ax-12 2172  ax-13 2372  ax-ext 2704
This theorem depends on definitions:  df-bi 206  df-an 398  df-or 847  df-tru 1545  df-ex 1783  df-nf 1787  df-sb 2069  df-clab 2711  df-cleq 2725  df-cdeq 3758
This theorem is referenced by: (None)
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