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Theorem dfalseu2 17189
Description: An "all some one" statement is equivalent to its universal part conjoined with the claim that exactly one 𝑥 satisfies both 𝜑 and 𝜓. In other words, given 𝑥(𝜑𝜓), requiring exactly one 𝑥 to satisfy 𝜑, which is what df-alseu 17174 requires, and requiring exactly one 𝑥 to satisfy (𝜑𝜓) come to the same thing. Read 𝜑 as "is a king" and 𝜓 as "is hungry": if every king is hungry, then "there is exactly one king" and "there is exactly one hungry king" say the same thing, so either of them, together with "every king is hungry", gives "the king is hungry".

The universal conjunct is what makes that work, and it cannot be dropped. ∃!𝑥(𝜑𝜓) on its own is strictly weaker than ∀∃!𝑥(𝜑𝜓), since it is satisfied when many things are 𝜑 and just one of those is 𝜓, as in a region with five kings exactly one of whom is hungry; see alseueu 17190 for the one direction that does hold without it. Uniqueness attaches to the antecedent, not to the conjunction. Russell's analysis of a definite description is built the same way: its uniqueness clause constrains the description predicate alone, while the predication is a separate conjunct. See his worked example of "the father of Charles II was executed", [Russell1905] p. 482. (Contributed by David A. Wheeler, 22-Jul-2026.)

Assertion
Ref Expression
dfalseu2 (∀∃!𝑥(𝜑𝜓) ↔ (∀𝑥(𝜑𝜓) ∧ ∃!𝑥(𝜑𝜓)))

Proof of Theorem dfalseu2
StepHypRef Expression
1 df-alseu 17174 . 2 (∀∃!𝑥(𝜑𝜓) ↔ (∀𝑥(𝜑𝜓) ∧ ∃!𝑥𝜑))
2 nfa1 1594 . . . 4 𝑥𝑥(𝜑𝜓)
3 sp 1564 . . . . 5 (∀𝑥(𝜑𝜓) → (𝜑𝜓))
4 pm4.71 393 . . . . 5 ((𝜑𝜓) ↔ (𝜑 ↔ (𝜑𝜓)))
53, 4sylib 122 . . . 4 (∀𝑥(𝜑𝜓) → (𝜑 ↔ (𝜑𝜓)))
62, 5eubid 2093 . . 3 (∀𝑥(𝜑𝜓) → (∃!𝑥𝜑 ↔ ∃!𝑥(𝜑𝜓)))
76pm5.32i 458 . 2 ((∀𝑥(𝜑𝜓) ∧ ∃!𝑥𝜑) ↔ (∀𝑥(𝜑𝜓) ∧ ∃!𝑥(𝜑𝜓)))
81, 7bitri 184 1 (∀∃!𝑥(𝜑𝜓) ↔ (∀𝑥(𝜑𝜓) ∧ ∃!𝑥(𝜑𝜓)))
Colors of variables:    wff set class
This proof depends on syntax axioms:  wi 4  wa 104  wb 105  wal 1400  ∃!weu 2086  ∀∃!walseu 17172
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1500  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-4 1563  ax-17 1579  ax-ial 1587
This proof depends on definitions:  df-bi 117  df-nf 1514  df-eu 2089  df-alseu 17174
This theorem is used by:  alseueu  17190
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