Theorem cdeqab1 3672

Description: Distribute conditional equality over abstraction. (Contributed by Mario Carneiro, 11-Aug-2016.)

Hypothesis

Ref	Expression
cdeqnot.1	⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
cdeqab1	⊢ CondEq(𝑥 = 𝑦 → {𝑥 ∣ 𝜑} = {𝑦 ∣ 𝜓})

Distinct variable groups:   𝜓,𝑥   𝜑,𝑦

Allowed substitution hints:   𝜑(𝑥)   𝜓(𝑦)

Proof of Theorem cdeqab1

Step	Hyp	Ref	Expression
1		nfv 1873	. . 3 ⊢ Ⅎ𝑦𝜑
2		nfv 1873	. . 3 ⊢ Ⅎ𝑥𝜓
3		cdeqnot.1	. . . 4 ⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
4	3	cdeqri 3666	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
5	1, 2, 4	cbvab 2911	. 2 ⊢ {𝑥 ∣ 𝜑} = {𝑦 ∣ 𝜓}
6	5	cdeqth 3667	1 ⊢ CondEq(𝑥 = 𝑦 → {𝑥 ∣ 𝜑} = {𝑦 ∣ 𝜓})

Syntax hints:   ↔ wb 198   = wceq 1507  {cab 2758  CondEqwcdeq 3663

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1758  ax-4 1772  ax-5 1869  ax-6 1928  ax-7 1965  ax-9 2059  ax-10 2079  ax-11 2093  ax-12 2106  ax-13 2301  ax-ext 2750

This theorem depends on definitions:  df-bi 199  df-an 388  df-or 834  df-tru 1510  df-ex 1743  df-nf 1747  df-sb 2016  df-clab 2759  df-cleq 2771  df-cdeq 3664

This theorem is referenced by: (None)