Theorem alseud 51105

Description: Introduction rule: "all some one" holds if the "for all" part holds and the antecedent has exactly one witness. This is the converse of alseu1d 51107 and alseu2d 51108 taken together. (Contributed by David A. Wheeler, 21-Jul-2026.)

Hypotheses

Ref	Expression
alseud.1	⊢ (𝜑 → ∀𝑥(𝜓 → 𝜒))
alseud.2	⊢ (𝜑 → ∃!𝑥𝜓)

Assertion

Ref	Expression
alseud	⊢ (𝜑 → ∀∃!𝑥(𝜓 → 𝜒))

Proof of Theorem alseud

Step	Hyp	Ref	Expression
1		alseud.1	. 2 ⊢ (𝜑 → ∀𝑥(𝜓 → 𝜒))
2		alseud.2	. 2 ⊢ (𝜑 → ∃!𝑥𝜓)
3		df-alseu 51100	. 2 ⊢ (∀∃!𝑥(𝜓 → 𝜒) ↔ (∀𝑥(𝜓 → 𝜒) ∧ ∃!𝑥𝜓))
4	1, 2, 3	sylanbrc 595	1 ⊢ (𝜑 → ∀∃!𝑥(𝜓 → 𝜒))

This proof depends on syntax axioms:   → wi 4  ∀wal 1567  ∃!weu 2593  ∀∃!walseu 51098

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-alseu 51100

This theorem is used by: (None)