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Theorem alsi-no-surprise 42351
 Description: Demonstrate that there is never a "surprise" when using the allsome quantifier, that is, it is never possible for the consequent to be both always true and always false. This uses the definition of df-alsi 42343; the proof itself builds on alimp-no-surprise 42336. For a contrast, see alimp-surprise 42335. (Contributed by David A. Wheeler, 27-Oct-2018.)
Assertion
Ref Expression
alsi-no-surprise ¬ (∀!𝑥(𝜑𝜓) ∧ ∀!𝑥(𝜑 → ¬ 𝜓))

Proof of Theorem alsi-no-surprise
StepHypRef Expression
1 alimp-no-surprise 42336 . 2 ¬ (∀𝑥(𝜑𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑)
2 df-alsi 42343 . . . 4 (∀!𝑥(𝜑𝜓) ↔ (∀𝑥(𝜑𝜓) ∧ ∃𝑥𝜑))
3 df-alsi 42343 . . . 4 (∀!𝑥(𝜑 → ¬ 𝜓) ↔ (∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑))
42, 3anbi12i 729 . . 3 ((∀!𝑥(𝜑𝜓) ∧ ∀!𝑥(𝜑 → ¬ 𝜓)) ↔ ((∀𝑥(𝜑𝜓) ∧ ∃𝑥𝜑) ∧ (∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑)))
5 anandi3r 1046 . . 3 ((∀𝑥(𝜑𝜓) ∧ ∃𝑥𝜑 ∧ ∀𝑥(𝜑 → ¬ 𝜓)) ↔ ((∀𝑥(𝜑𝜓) ∧ ∃𝑥𝜑) ∧ (∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑)))
6 3ancomb 1040 . . 3 ((∀𝑥(𝜑𝜓) ∧ ∃𝑥𝜑 ∧ ∀𝑥(𝜑 → ¬ 𝜓)) ↔ (∀𝑥(𝜑𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑))
74, 5, 63bitr2i 287 . 2 ((∀!𝑥(𝜑𝜓) ∧ ∀!𝑥(𝜑 → ¬ 𝜓)) ↔ (∀𝑥(𝜑𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑))
81, 7mtbir 312 1 ¬ (∀!𝑥(𝜑𝜓) ∧ ∀!𝑥(𝜑 → ¬ 𝜓))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 383   ∧ w3a 1031  ∀wal 1473  ∃wex 1695  ∀!walsi 42341 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728 This theorem depends on definitions:  df-bi 196  df-an 385  df-3an 1033  df-ex 1696  df-alsi 42343 This theorem is referenced by: (None)
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