Theorem cdeqeq 2950

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 2941	. . 3 ⊢ (𝑥 = 𝑦 → 𝐴 = 𝐵)
3		cdeqeq.2	. . . 4 ⊢ CondEq(𝑥 = 𝑦 → 𝐶 = 𝐷)
4	3	cdeqri 2941	. . 3 ⊢ (𝑥 = 𝑦 → 𝐶 = 𝐷)
5	2, 4	eqeq12d 2185	. 2 ⊢ (𝑥 = 𝑦 → (𝐴 = 𝐶 ↔ 𝐵 = 𝐷))
6	5	cdeqi 2940	1 ⊢ CondEq(𝑥 = 𝑦 → (𝐴 = 𝐶 ↔ 𝐵 = 𝐷))

Syntax hints:   ↔ wb 104   = wceq 1348  CondEqwcdeq 2938

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1440  ax-gen 1442  ax-4 1503  ax-17 1519  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-cleq 2163  df-cdeq 2939

This theorem is referenced by: (None)