Theorem alseud 50661

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 50663 and alseu2d 50664 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 50656	. 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 1568  ∃!weu 2598  ∀∃!walseu 50654

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 50656

This theorem is used by: (None)