Theorem rals1d 50630

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.) (Revised by David A. Wheeler, 12-Jul-2026.)

Hypothesis

Ref	Expression
rals1d.1	⊢ (𝜑 → ∀∃𝑥 ∈ 𝐴(𝜓 → 𝜒))

Assertion

Ref	Expression
rals1d	⊢ (𝜑 → ∀𝑥 ∈ 𝐴 (𝜓 → 𝜒))

Proof of Theorem rals1d

Step	Hyp	Ref	Expression
1		rals1d.1	. . 3 ⊢ (𝜑 → ∀∃𝑥 ∈ 𝐴(𝜓 → 𝜒))
2		df-rals 50624	. . 3 ⊢ (∀∃𝑥 ∈ 𝐴(𝜓 → 𝜒) ↔ (∀𝑥 ∈ 𝐴 (𝜓 → 𝜒) ∧ ∃𝑥 ∈ 𝐴 𝜓))
3	1, 2	sylib 221	. 2 ⊢ (𝜑 → (∀𝑥 ∈ 𝐴 (𝜓 → 𝜒) ∧ ∃𝑥 ∈ 𝐴 𝜓))
4	3	simpld 500	1 ⊢ (𝜑 → ∀𝑥 ∈ 𝐴 (𝜓 → 𝜒))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401  ∀wral 3081  ∃wrex 3091  ∀∃wrals 50622

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-an 402  df-rals 50624

This theorem is used by: (None)