Theorem alseu-no-surprise 51117

Description: 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 51085 by alseuals 51103. For a contrast, see alimp-surprise 51059. (Contributed by David A. Wheeler, 21-Jul-2026.)

Assertion

Ref	Expression
alseu-no-surprise	⊢ ¬ (∀∃!𝑥(𝜑 → 𝜓) ∧ ∀∃!𝑥(𝜑 → ¬ 𝜓))

Proof of Theorem alseu-no-surprise

Step	Hyp	Ref	Expression
1		als-no-surprise 51085	. 2 ⊢ ¬ (∀∃𝑥(𝜑 → 𝜓) ∧ ∀∃𝑥(𝜑 → ¬ 𝜓))
2		alseuals 51103	. . 3 ⊢ (∀∃!𝑥(𝜑 → 𝜓) → ∀∃𝑥(𝜑 → 𝜓))
3		alseuals 51103	. . 3 ⊢ (∀∃!𝑥(𝜑 → ¬ 𝜓) → ∀∃𝑥(𝜑 → ¬ 𝜓))
4	2, 3	anim12i 625	. 2 ⊢ ((∀∃!𝑥(𝜑 → 𝜓) ∧ ∀∃!𝑥(𝜑 → ¬ 𝜓)) → (∀∃𝑥(𝜑 → 𝜓) ∧ ∀∃𝑥(𝜑 → ¬ 𝜓)))
5	1, 4	mto 200	1 ⊢ ¬ (∀∃!𝑥(𝜑 → 𝜓) ∧ ∀∃!𝑥(𝜑 → ¬ 𝜓))

This proof depends on syntax axioms:  ¬ wn 3   ∧ wa 401  ∀∃wals 51065  ∀∃!walseu 51098

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841

This proof depends on definitions:  df-bi 210  df-an 402  df-3an 1105  df-ex 1812  df-eu 2594  df-als 51067  df-alseu 51100

This theorem is used by: (None)