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Theorem cdeqab1 2821
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
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 cdeqnot.1 . . . 4 CondEq(𝑥 = 𝑦 → (𝜑𝜓))
21cdeqri 2815 . . 3 (𝑥 = 𝑦 → (𝜑𝜓))
32cbvabv 2208 . 2 {𝑥𝜑} = {𝑦𝜓}
43cdeqth 2816 1 CondEq(𝑥 = 𝑦 → {𝑥𝜑} = {𝑦𝜓})
Colors of variables: wff set class
Syntax hints:  wb 103   = wceq 1287  {cab 2071  CondEqwcdeq 2812
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 663  ax-5 1379  ax-7 1380  ax-gen 1381  ax-ie1 1425  ax-ie2 1426  ax-8 1438  ax-10 1439  ax-11 1440  ax-i12 1441  ax-bndl 1442  ax-4 1443  ax-17 1462  ax-i9 1466  ax-ial 1470  ax-i5r 1471  ax-ext 2067
This theorem depends on definitions:  df-bi 115  df-nf 1393  df-sb 1690  df-clab 2072  df-cleq 2078  df-cdeq 2813
This theorem is referenced by: (None)
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