Theorem cdeqeq 2810

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 2801	. . 3 ⊢ (𝑥 = 𝑦 → 𝐴 = 𝐵)
3		cdeqeq.2	. . . 4 ⊢ CondEq(𝑥 = 𝑦 → 𝐶 = 𝐷)
4	3	cdeqri 2801	. . 3 ⊢ (𝑥 = 𝑦 → 𝐶 = 𝐷)
5	2, 4	eqeq12d 2095	. 2 ⊢ (𝑥 = 𝑦 → (𝐴 = 𝐶 ↔ 𝐵 = 𝐷))
6	5	cdeqi 2800	1 ⊢ CondEq(𝑥 = 𝑦 → (𝐴 = 𝐶 ↔ 𝐵 = 𝐷))

Syntax hints:   ↔ wb 103   = wceq 1284  CondEqwcdeq 2798

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1376  ax-gen 1378  ax-4 1440  ax-17 1459  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-cleq 2074  df-cdeq 2799

This theorem is referenced by: (None)