Theorem cdeqel 3798

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 3788	. . 3 ⊢ (𝑥 = 𝑦 → 𝐴 = 𝐵)
3		cdeqeq.2	. . . 4 ⊢ CondEq(𝑥 = 𝑦 → 𝐶 = 𝐷)
4	3	cdeqri 3788	. . 3 ⊢ (𝑥 = 𝑦 → 𝐶 = 𝐷)
5	2, 4	eleq12d 2838	. 2 ⊢ (𝑥 = 𝑦 → (𝐴 ∈ 𝐶 ↔ 𝐵 ∈ 𝐷))
6	5	cdeqi 3787	1 ⊢ CondEq(𝑥 = 𝑦 → (𝐴 ∈ 𝐶 ↔ 𝐵 ∈ 𝐷))

Syntax hints:   ↔ wb 206   = wceq 1537   ∈ wcel 2108  CondEqwcdeq 3785

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1778  df-cleq 2732  df-clel 2819  df-cdeq 3786

This theorem is referenced by:  nfccdeq  3800