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Theorem cdeqab1 3394
 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 3388 . . 3 (𝑥 = 𝑦 → (𝜑𝜓))
32cbvabv 2734 . 2 {𝑥𝜑} = {𝑦𝜓}
43cdeqth 3389 1 CondEq(𝑥 = 𝑦 → {𝑥𝜑} = {𝑦𝜓})
 Colors of variables: wff setvar class Syntax hints:   ↔ wb 195   = wceq 1475  {cab 2596  CondEqwcdeq 3385 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590 This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-cdeq 3386 This theorem is referenced by: (None)
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