Theorem cdeqeq 3768

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 3759	. . 3 ⊢ (𝑥 = 𝑦 → 𝐴 = 𝐵)
3		cdeqeq.2	. . . 4 ⊢ CondEq(𝑥 = 𝑦 → 𝐶 = 𝐷)
4	3	cdeqri 3759	. . 3 ⊢ (𝑥 = 𝑦 → 𝐶 = 𝐷)
5	2, 4	eqeq12d 2839	. 2 ⊢ (𝑥 = 𝑦 → (𝐴 = 𝐶 ↔ 𝐵 = 𝐷))
6	5	cdeqi 3758	1 ⊢ CondEq(𝑥 = 𝑦 → (𝐴 = 𝐶 ↔ 𝐵 = 𝐷))

Syntax hints:   ↔ wb 208   = wceq 1537  CondEqwcdeq 3756

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-9 2124  ax-ext 2795

This theorem depends on definitions:  df-bi 209  df-an 399  df-ex 1781  df-cleq 2816  df-cdeq 3757

This theorem is referenced by: (None)