Metamath Proof Explorer


Theorem ralseu2d

Description: Deduction rule: Given "all some one" applied to a class, you can extract the "exactly one" part. Note that the witness must satisfy the antecedent ps , not merely be a member of A . (Contributed by David A. Wheeler, 21-Jul-2026)

Ref Expression
Hypothesis ralseu2d.1 ⊢ φ → ∀∃! x ∈ A ψ → χ
Assertion ralseu2d ⊢ φ → ∃! x ∈ A ψ

Proof

Step Hyp Ref Expression
1 ralseu2d.1 ⊢ φ → ∀∃! x ∈ A ψ → χ
2 df-ralseu ⊢ ∀∃! 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 simprd ⊢ φ → ∃! x ∈ A ψ