Metamath Proof Explorer


Theorem rals1d

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)

Ref Expression
Hypothesis rals1d.1 ⊢ φ → ∀∃ x ∈ A ψ → χ
Assertion rals1d ⊢ φ → ∀ x ∈ A ψ → χ

Proof

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