Theorem alseuals 50602

Description: "All some one" implies "all some": requiring exactly one witness is stronger than requiring at least one. Any consequence of an allsome statement is therefore a consequence of the corresponding "all some one" statement, which is how alseu-no-surprise 50616 is proved. (Contributed by David A. Wheeler, 21-Jul-2026.)

Assertion

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

Proof of Theorem alseuals

Step	Hyp	Ref	Expression
1		euex 2605	. . 3 ⊢ (∃!𝑥𝜑 → ∃𝑥𝜑)
2	1	anim2i 628	. 2 ⊢ ((∀𝑥(𝜑 → 𝜓) ∧ ∃!𝑥𝜑) → (∀𝑥(𝜑 → 𝜓) ∧ ∃𝑥𝜑))
3		df-alseu 50599	. 2 ⊢ (∀∃!𝑥(𝜑 → 𝜓) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∃!𝑥𝜑))
4		df-als 50566	. 2 ⊢ (∀∃𝑥(𝜑 → 𝜓) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∃𝑥𝜑))
5	2, 3, 4	3imtr4i 295	1 ⊢ (∀∃!𝑥(𝜑 → 𝜓) → ∀∃𝑥(𝜑 → 𝜓))

Syntax hints:   → wi 4   ∧ wa 400  ∀wal 1568  ∃wex 1809  ∃!weu 2596  ∀∃wals 50564  ∀∃!walseu 50597

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 210  df-an 401  df-eu 2597  df-als 50566  df-alseu 50599

This theorem is referenced by:  alseu-no-surprise  50616