Theorem cdeqab1 3729

Description: Distribute conditional equality over abstraction. Usage of this theorem is discouraged because it depends on ax-13 2401. (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 1946	. . 3 ⊢ Ⅎ𝑦𝜑
2		nfv 1946	. . 3 ⊢ Ⅎ𝑥𝜓
3		cdeqnot.1	. . . 4 ⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
4	3	cdeqri 3723	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
5	1, 2, 4	cbvab 2832	. 2 ⊢ {𝑥 ∣ 𝜑} = {𝑦 ∣ 𝜓}
6	5	cdeqth 3724	1 ⊢ CondEq(𝑥 = 𝑦 → {𝑥 ∣ 𝜑} = {𝑦 ∣ 𝜓})

This proof depends on syntax axioms:   ↔ wb 209   = wceq 1569  {cab 2738  CondEqwcdeq 3720

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-9 2154  ax-10 2177  ax-11 2193  ax-12 2212  ax-13 2401  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-cdeq 3721

This theorem is used by: (None)