Theorem cdeqeq 3397

Description: Distribute conditional equality over equality. (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
cdeqeq	⊢ CondEq(𝑥 = 𝑦 → (𝐴 = 𝐶 ↔ 𝐵 = 𝐷))

Proof of Theorem cdeqeq

Step	Hyp	Ref	Expression
1		cdeqeq.1	. . . 4 ⊢ CondEq(𝑥 = 𝑦 → 𝐴 = 𝐵)
2	1	cdeqri 3388	. . 3 ⊢ (𝑥 = 𝑦 → 𝐴 = 𝐵)
3		cdeqeq.2	. . . 4 ⊢ CondEq(𝑥 = 𝑦 → 𝐶 = 𝐷)
4	3	cdeqri 3388	. . 3 ⊢ (𝑥 = 𝑦 → 𝐶 = 𝐷)
5	2, 4	eqeq12d 2625	. 2 ⊢ (𝑥 = 𝑦 → (𝐴 = 𝐶 ↔ 𝐵 = 𝐷))
6	5	cdeqi 3387	1 ⊢ CondEq(𝑥 = 𝑦 → (𝐴 = 𝐶 ↔ 𝐵 = 𝐷))

Syntax hints:   ↔ wb 195   = wceq 1475  CondEqwcdeq 3385

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-ext 2590

This theorem depends on definitions:  df-bi 196  df-an 385  df-cleq 2603  df-cdeq 3386

This theorem is referenced by: (None)