Theorem cdeqal1 3736

Description: Distribute conditional equality over quantification. Usage of this theorem is discouraged because it depends on ax-13 2406. (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 3731	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
3	2	cbvalv 2434	. 2 ⊢ (∀𝑥𝜑 ↔ ∀𝑦𝜓)
4	3	cdeqth 3732	1 ⊢ CondEq(𝑥 = 𝑦 → (∀𝑥𝜑 ↔ ∀𝑦𝜓))

This proof depends on syntax axioms:   ↔ wb 209  ∀wal 1568  CondEqwcdeq 3728

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  ax-6 2000  ax-7 2041  ax-11 2195  ax-12 2216  ax-13 2406

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-nf 1817  df-cdeq 3729

This theorem is used by: (None)