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Theorem alimp-no-surprise 43048
 Description: There is no "surprise" in a for-all with implication if there exists a value where the antecedent is true. This is one way to prevent for-all with implication from allowing anything. For a contrast, see alimp-surprise 43047. The allsome quantifier also counters this problem, see df-alsi 43055. (Contributed by David A. Wheeler, 27-Oct-2018.)
Assertion
Ref Expression
alimp-no-surprise ¬ (∀𝑥(𝜑𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑)

Proof of Theorem alimp-no-surprise
StepHypRef Expression
1 pm4.82 1005 . . . . 5 (((𝜑𝜓) ∧ (𝜑 → ¬ 𝜓)) ↔ ¬ 𝜑)
21albii 1894 . . . 4 (∀𝑥((𝜑𝜓) ∧ (𝜑 → ¬ 𝜓)) ↔ ∀𝑥 ¬ 𝜑)
3 alnex 1853 . . . 4 (∀𝑥 ¬ 𝜑 ↔ ¬ ∃𝑥𝜑)
42, 3sylbb 209 . . 3 (∀𝑥((𝜑𝜓) ∧ (𝜑 → ¬ 𝜓)) → ¬ ∃𝑥𝜑)
5 imnan 386 . . 3 ((∀𝑥((𝜑𝜓) ∧ (𝜑 → ¬ 𝜓)) → ¬ ∃𝑥𝜑) ↔ ¬ (∀𝑥((𝜑𝜓) ∧ (𝜑 → ¬ 𝜓)) ∧ ∃𝑥𝜑))
64, 5mpbi 220 . 2 ¬ (∀𝑥((𝜑𝜓) ∧ (𝜑 → ¬ 𝜓)) ∧ ∃𝑥𝜑)
7 19.26 1948 . . . 4 (∀𝑥((𝜑𝜓) ∧ (𝜑 → ¬ 𝜓)) ↔ (∀𝑥(𝜑𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓)))
87anbi2ci 603 . . 3 ((∀𝑥((𝜑𝜓) ∧ (𝜑 → ¬ 𝜓)) ∧ ∃𝑥𝜑) ↔ (∃𝑥𝜑 ∧ (∀𝑥(𝜑𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓))))
9 3anass 1079 . . 3 ((∃𝑥𝜑 ∧ ∀𝑥(𝜑𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓)) ↔ (∃𝑥𝜑 ∧ (∀𝑥(𝜑𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓))))
10 3anrot 1085 . . 3 ((∃𝑥𝜑 ∧ ∀𝑥(𝜑𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓)) ↔ (∀𝑥(𝜑𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑))
118, 9, 103bitr2i 288 . 2 ((∀𝑥((𝜑𝜓) ∧ (𝜑 → ¬ 𝜓)) ∧ ∃𝑥𝜑) ↔ (∀𝑥(𝜑𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑))
126, 11mtbi 311 1 ¬ (∀𝑥(𝜑𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑)
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 382   ∧ w3a 1070  ∀wal 1628  ∃wex 1851 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1869  ax-4 1884 This theorem depends on definitions:  df-bi 197  df-an 383  df-3an 1072  df-ex 1852 This theorem is referenced by:  alsi-no-surprise  43063
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