Theorem cdeqal1 3766

Description: Distribute conditional equality over quantification. Usage of this theorem is discouraged because it depends on ax-13 2372. (Contributed by Mario Carneiro, 11-Aug-2016.) (New usage is discouraged.)

Hypothesis

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

Assertion

Ref	Expression
cdeqal1	⊢ CondEq(𝑥 = 𝑦 → (∀𝑥𝜑 ↔ ∀𝑦𝜓))

Distinct variable groups:   𝜓,𝑥   𝜑,𝑦

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

Proof of Theorem cdeqal1

Step	Hyp	Ref	Expression
1		cdeqnot.1	. . . 4 ⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
2	1	cdeqri 3761	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
3	2	cbvalv 2400	. 2 ⊢ (∀𝑥𝜑 ↔ ∀𝑦𝜓)
4	3	cdeqth 3762	1 ⊢ CondEq(𝑥 = 𝑦 → (∀𝑥𝜑 ↔ ∀𝑦𝜓))

Syntax hints:   ↔ wb 205  ∀wal 1540  CondEqwcdeq 3758

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-11 2155  ax-12 2172  ax-13 2372

This theorem depends on definitions:  df-bi 206  df-an 398  df-ex 1783  df-nf 1787  df-cdeq 3759

This theorem is referenced by: (None)