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 )