Theorem alsc1d 13959

Description: Deduction rule: Given "all some" applied to a class, you can extract the "for all" part. (Contributed by David A. Wheeler, 20-Oct-2018.)

Hypothesis

Ref	Expression
alsc1d.1	⊢ (𝜑 → ∀!𝑥 ∈ 𝐴𝜓)

Assertion

Ref	Expression
alsc1d	⊢ (𝜑 → ∀𝑥 ∈ 𝐴 𝜓)

Proof of Theorem alsc1d

Step	Hyp	Ref	Expression
1		alsc1d.1	. . 3 ⊢ (𝜑 → ∀!𝑥 ∈ 𝐴𝜓)
2		df-alsc 13955	. . 3 ⊢ (∀!𝑥 ∈ 𝐴𝜓 ↔ (∀𝑥 ∈ 𝐴 𝜓 ∧ ∃𝑥 𝑥 ∈ 𝐴))
3	1, 2	sylib 121	. 2 ⊢ (𝜑 → (∀𝑥 ∈ 𝐴 𝜓 ∧ ∃𝑥 𝑥 ∈ 𝐴))
4	3	simpld 111	1 ⊢ (𝜑 → ∀𝑥 ∈ 𝐴 𝜓)

Syntax hints:   → wi 4   ∧ wa 103  ∃wex 1480   ∈ wcel 2136  ∀wral 2444  ∀!walsc 13953

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105

This theorem depends on definitions:  df-bi 116  df-alsc 13955

This theorem is referenced by: (None)