Theorem cdeqal1 2993

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 2988	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
3	2	cbvalv 1942	. 2 ⊢ (∀𝑥𝜑 ↔ ∀𝑦𝜓)
4	3	cdeqth 2989	1 ⊢ CondEq(𝑥 = 𝑦 → (∀𝑥𝜑 ↔ ∀𝑦𝜓))

Syntax hints:   ↔ wb 105  ∀wal 1371  CondEqwcdeq 2985

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558

This theorem depends on definitions:  df-bi 117  df-nf 1485  df-cdeq 2986

This theorem is referenced by: (None)