Theorem cdeqal 3735

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 3732	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
3	2	albidv 1953	. 2 ⊢ (𝑥 = 𝑦 → (∀𝑧𝜑 ↔ ∀𝑧𝜓))
4	3	cdeqi 3731	1 ⊢ CondEq(𝑥 = 𝑦 → (∀𝑧𝜑 ↔ ∀𝑧𝜓))

This proof depends on syntax axioms:   ↔ wb 209  ∀wal 1568  CondEqwcdeq 3729

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943

This proof depends on definitions:  df-bi 210  df-cdeq 3730

This theorem is used by: (None)