P. 175 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 17401-17403   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	dfalseu2 17401	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(𝜑 → 𝜓), requiring exactly one 𝑥 to satisfy 𝜑, which is what df-alseu 17386 requires, and requiring exactly one 𝑥 to satisfy (𝜑 ∧ 𝜓) 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(𝜑 ∧ 𝜓) on its own is strictly weaker than ∀∃!𝑥(𝜑 → 𝜓), 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 17402 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.)
⊢ (∀∃!𝑥(𝜑 → 𝜓) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∃!𝑥(𝜑 ∧ 𝜓)))
Theorem	alseueu 17402	"The 𝜑 is 𝜓 " implies that exactly one thing is both 𝜑 and 𝜓. This is the half of dfalseu2 17401 that drops the universal conjunct; it does not reverse, so ∃!𝑥(𝜑 ∧ 𝜓) cannot be used in place of an "all some one" statement. (Contributed by David A. Wheeler, 22-Jul-2026.)
⊢ (∀∃!𝑥(𝜑 → 𝜓) → ∃!𝑥(𝜑 ∧ 𝜓))
Theorem	alseu-no-surprise 17403	Demonstrate that there is never a "surprise" when using the "all some one" quantifier, that is, it is never possible for the consequent to be both always true and always false. This follows from als-no-surprise 17371 by alseuals 17389. See als-no-surprise 17371 for why ordinary "for all" with implication has no such property. (Contributed by David A. Wheeler, 22-Jul-2026.)
⊢ ¬ (∀∃!𝑥(𝜑 → 𝜓) ∧ ∀∃!𝑥(𝜑 → ¬ 𝜓))

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