Theorem cdeqim 2902

Description: Distribute conditional equality over implication. (Contributed by Mario Carneiro, 11-Aug-2016.)

Hypotheses

Ref	Expression
cdeqnot.1	⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
cdeqim.1	⊢ CondEq(𝑥 = 𝑦 → (𝜒 ↔ 𝜃))

Assertion

Ref	Expression
cdeqim	⊢ CondEq(𝑥 = 𝑦 → ((𝜑 → 𝜒) ↔ (𝜓 → 𝜃)))

Proof of Theorem cdeqim

Step	Hyp	Ref	Expression
1		cdeqnot.1	. . . 4 ⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
2	1	cdeqri 2895	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
3		cdeqim.1	. . . 4 ⊢ CondEq(𝑥 = 𝑦 → (𝜒 ↔ 𝜃))
4	3	cdeqri 2895	. . 3 ⊢ (𝑥 = 𝑦 → (𝜒 ↔ 𝜃))
5	2, 4	imbi12d 233	. 2 ⊢ (𝑥 = 𝑦 → ((𝜑 → 𝜒) ↔ (𝜓 → 𝜃)))
6	5	cdeqi 2894	1 ⊢ CondEq(𝑥 = 𝑦 → ((𝜑 → 𝜒) ↔ (𝜓 → 𝜃)))

Syntax hints:   → wi 4   ↔ wb 104  CondEqwcdeq 2892

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107

This theorem depends on definitions:  df-bi 116  df-cdeq 2893

This theorem is referenced by: (None)