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Theorem cdeqab 2927
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 2923 . . 3 (𝑥 = 𝑦 → (𝜑𝜓))
32abbidv 2275 . 2 (𝑥 = 𝑦 → {𝑧𝜑} = {𝑧𝜓})
43cdeqi 2922 1 CondEq(𝑥 = 𝑦 → {𝑧𝜑} = {𝑧𝜓})
Colors of variables: wff set class
Syntax hints:  wb 104   = wceq 1335  {cab 2143  CondEqwcdeq 2920
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1427  ax-7 1428  ax-gen 1429  ax-ie1 1473  ax-ie2 1474  ax-8 1484  ax-11 1486  ax-4 1490  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2139
This theorem depends on definitions:  df-bi 116  df-tru 1338  df-nf 1441  df-sb 1743  df-clab 2144  df-cleq 2150  df-cdeq 2921
This theorem is referenced by: (None)
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