Theorem alseuals 17139

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 17153 is proved. (Contributed by David A. Wheeler, 22-Jul-2026.)

Assertion

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

Proof of Theorem alseuals

Step	Hyp	Ref	Expression
1		euex 2116	. . 3 ⊢ (∃!𝑥𝜑 → ∃𝑥𝜑)
2	1	anim2i 342	. 2 ⊢ ((∀𝑥(𝜑 → 𝜓) ∧ ∃!𝑥𝜑) → (∀𝑥(𝜑 → 𝜓) ∧ ∃𝑥𝜑))
3		df-alseu 17136	. 2 ⊢ (∀∃!𝑥(𝜑 → 𝜓) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∃!𝑥𝜑))
4		df-als 17102	. 2 ⊢ (∀∃𝑥(𝜑 → 𝜓) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∃𝑥𝜑))
5	2, 3, 4	3imtr4i 201	1 ⊢ (∀∃!𝑥(𝜑 → 𝜓) → ∀∃𝑥(𝜑 → 𝜓))

Syntax hints:   → wi 4   ∧ wa 104  ∀wal 1400  ∃wex 1545  ∃!weu 2086  ∀∃wals 17100  ∀∃!walseu 17134

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588

This theorem depends on definitions:  df-bi 117  df-nf 1514  df-sb 1816  df-eu 2089  df-als 17102  df-alseu 17136

This theorem is referenced by:  alseu-no-surprise  17153