Theorem cdeqab1 3744

Description: Distribute conditional equality over abstraction. Usage of this theorem is discouraged because it depends on ax-13 2410. (Contributed by Mario Carneiro, 11-Aug-2016.) (New usage is discouraged.)

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 1941	. . 3 ⊢ Ⅎ𝑦𝜑
2		nfv 1941	. . 3 ⊢ Ⅎ𝑥𝜓
3		cdeqnot.1	. . . 4 ⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
4	3	cdeqri 3738	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
5	1, 2, 4	cbvab 2841	. 2 ⊢ {𝑥 ∣ 𝜑} = {𝑦 ∣ 𝜓}
6	5	cdeqth 3739	1 ⊢ CondEq(𝑥 = 𝑦 → {𝑥 ∣ 𝜑} = {𝑦 ∣ 𝜓})

Syntax hints:   ↔ wb 209   = wceq 1567  {cab 2747  CondEqwcdeq 3735

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-9 2159  ax-10 2182  ax-11 2198  ax-12 2219  ax-13 2410  ax-ext 2741

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1570  df-ex 1807  df-nf 1811  df-sb 2098  df-clab 2748  df-cleq 2761  df-cdeq 3736

This theorem is referenced by: (None)