Theorem cdeqab1 3768

Description: Distribute conditional equality over abstraction. Usage of this theorem is discouraged because it depends on ax-13 2370. (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 1916	. . 3 ⊢ Ⅎ𝑦𝜑
2		nfv 1916	. . 3 ⊢ Ⅎ𝑥𝜓
3		cdeqnot.1	. . . 4 ⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
4	3	cdeqri 3762	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
5	1, 2, 4	cbvab 2807	. 2 ⊢ {𝑥 ∣ 𝜑} = {𝑦 ∣ 𝜓}
6	5	cdeqth 3763	1 ⊢ CondEq(𝑥 = 𝑦 → {𝑥 ∣ 𝜑} = {𝑦 ∣ 𝜓})

Syntax hints:   ↔ wb 205   = wceq 1540  {cab 2708  CondEqwcdeq 3759

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1912  ax-6 1970  ax-7 2010  ax-9 2115  ax-10 2136  ax-11 2153  ax-12 2170  ax-13 2370  ax-ext 2702

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-tru 1543  df-ex 1781  df-nf 1785  df-sb 2067  df-clab 2709  df-cleq 2723  df-cdeq 3760

This theorem is referenced by: (None)