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Theorem cdeqab1 2756
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 2750 . . 3 (𝑥 = 𝑦 → (𝜑𝜓))
32cbvabv 2161 . 2 {𝑥𝜑} = {𝑦𝜓}
43cdeqth 2751 1 CondEq(𝑥 = 𝑦 → {𝑥𝜑} = {𝑦𝜓})
Colors of variables: wff set class
Syntax hints:  wb 98   = wceq 1243  {cab 2026  CondEqwcdeq 2747
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022
This theorem depends on definitions:  df-bi 110  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-cdeq 2748
This theorem is referenced by: (None)
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