Theorem cdeqal 3699

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

Hypothesis

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

Assertion

Ref	Expression
cdeqal	⊢ CondEq(𝑥 = 𝑦 → (∀𝑧𝜑 ↔ ∀𝑧𝜓))

Distinct variable groups:   𝑥,𝑧   𝑦,𝑧

Allowed substitution hints:   𝜑(𝑥,𝑦,𝑧)   𝜓(𝑥,𝑦,𝑧)

Proof of Theorem cdeqal

Step	Hyp	Ref	Expression
1		cdeqnot.1	. . . 4 ⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
2	1	cdeqri 3696	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
3	2	albidv 1924	. 2 ⊢ (𝑥 = 𝑦 → (∀𝑧𝜑 ↔ ∀𝑧𝜓))
4	3	cdeqi 3695	1 ⊢ CondEq(𝑥 = 𝑦 → (∀𝑧𝜑 ↔ ∀𝑧𝜓))

Syntax hints:   ↔ wb 205  ∀wal 1537  CondEqwcdeq 3693

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914

This theorem depends on definitions:  df-bi 206  df-cdeq 3694

This theorem is referenced by: (None)