Theorem cdeqal1 2806

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
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 2801	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
3	2	cbvalv 1835	. 2 ⊢ (∀𝑥𝜑 ↔ ∀𝑦𝜓)
4	3	cdeqth 2802	1 ⊢ CondEq(𝑥 = 𝑦 → (∀𝑥𝜑 ↔ ∀𝑦𝜓))

Syntax hints:   ↔ wb 103  ∀wal 1282  CondEqwcdeq 2798

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467

This theorem depends on definitions:  df-bi 115  df-nf 1390  df-cdeq 2799

This theorem is referenced by: (None)