Theorem cdeqel 3431

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

Hypotheses

Ref	Expression
cdeqeq.1	⊢ CondEq(𝑥 = 𝑦 → 𝐴 = 𝐵)
cdeqeq.2	⊢ CondEq(𝑥 = 𝑦 → 𝐶 = 𝐷)

Assertion

Ref	Expression
cdeqel	⊢ CondEq(𝑥 = 𝑦 → (𝐴 ∈ 𝐶 ↔ 𝐵 ∈ 𝐷))

Proof of Theorem cdeqel

Step	Hyp	Ref	Expression
1		cdeqeq.1	. . . 4 ⊢ CondEq(𝑥 = 𝑦 → 𝐴 = 𝐵)
2	1	cdeqri 3421	. . 3 ⊢ (𝑥 = 𝑦 → 𝐴 = 𝐵)
3		cdeqeq.2	. . . 4 ⊢ CondEq(𝑥 = 𝑦 → 𝐶 = 𝐷)
4	3	cdeqri 3421	. . 3 ⊢ (𝑥 = 𝑦 → 𝐶 = 𝐷)
5	2, 4	eleq12d 2695	. 2 ⊢ (𝑥 = 𝑦 → (𝐴 ∈ 𝐶 ↔ 𝐵 ∈ 𝐷))
6	5	cdeqi 3420	1 ⊢ CondEq(𝑥 = 𝑦 → (𝐴 ∈ 𝐶 ↔ 𝐵 ∈ 𝐷))

Syntax hints:   ↔ wb 196   = wceq 1483   ∈ wcel 1990  CondEqwcdeq 3418

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-an 386  df-ex 1705  df-cleq 2615  df-clel 2618  df-cdeq 3419

This theorem is referenced by:  nfccdeq  3433