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Theorem cdeqab1 2778
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 2772 . . 3 (𝑥 = 𝑦 → (𝜑𝜓))
32cbvabv 2177 . 2 {𝑥𝜑} = {𝑦𝜓}
43cdeqth 2773 1 CondEq(𝑥 = 𝑦 → {𝑥𝜑} = {𝑦𝜓})
Colors of variables: wff set class
Syntax hints:  wb 102   = wceq 1259  {cab 2042  CondEqwcdeq 2769
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-io 640  ax-5 1352  ax-7 1353  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-8 1411  ax-10 1412  ax-11 1413  ax-i12 1414  ax-bndl 1415  ax-4 1416  ax-17 1435  ax-i9 1439  ax-ial 1443  ax-i5r 1444  ax-ext 2038
This theorem depends on definitions:  df-bi 114  df-nf 1366  df-sb 1662  df-clab 2043  df-cleq 2049  df-cdeq 2770
This theorem is referenced by: (None)
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