Theorem cdeqel 2985

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 2975	. . 3 ⊢ (𝑥 = 𝑦 → 𝐴 = 𝐵)
3		cdeqeq.2	. . . 4 ⊢ CondEq(𝑥 = 𝑦 → 𝐶 = 𝐷)
4	3	cdeqri 2975	. . 3 ⊢ (𝑥 = 𝑦 → 𝐶 = 𝐷)
5	2, 4	eleq12d 2267	. 2 ⊢ (𝑥 = 𝑦 → (𝐴 ∈ 𝐶 ↔ 𝐵 ∈ 𝐷))
6	5	cdeqi 2974	1 ⊢ CondEq(𝑥 = 𝑦 → (𝐴 ∈ 𝐶 ↔ 𝐵 ∈ 𝐷))

Syntax hints:   ↔ wb 105   = wceq 1364   ∈ wcel 2167  CondEqwcdeq 2972

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1461  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-4 1524  ax-17 1540  ax-ial 1548  ax-ext 2178

This theorem depends on definitions:  df-bi 117  df-cleq 2189  df-clel 2192  df-cdeq 2973

This theorem is referenced by:  nfccdeq  2987