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Theorem alimp-no-surprise 41830
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 41829. The allsome quantifier also counters this problem, see df-alsi 41837. (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 pm4.82 968 . . . . 5 (((𝜑𝜓) ∧ (𝜑 → ¬ 𝜓)) ↔ ¬ 𝜑)
21albii 1744 . . . 4 (∀𝑥((𝜑𝜓) ∧ (𝜑 → ¬ 𝜓)) ↔ ∀𝑥 ¬ 𝜑)
3 alnex 1703 . . . 4 (∀𝑥 ¬ 𝜑 ↔ ¬ ∃𝑥𝜑)
42, 3sylbb 209 . . 3 (∀𝑥((𝜑𝜓) ∧ (𝜑 → ¬ 𝜓)) → ¬ ∃𝑥𝜑)
5 imnan 438 . . 3 ((∀𝑥((𝜑𝜓) ∧ (𝜑 → ¬ 𝜓)) → ¬ ∃𝑥𝜑) ↔ ¬ (∀𝑥((𝜑𝜓) ∧ (𝜑 → ¬ 𝜓)) ∧ ∃𝑥𝜑))
64, 5mpbi 220 . 2 ¬ (∀𝑥((𝜑𝜓) ∧ (𝜑 → ¬ 𝜓)) ∧ ∃𝑥𝜑)
7 19.26 1795 . . . 4 (∀𝑥((𝜑𝜓) ∧ (𝜑 → ¬ 𝜓)) ↔ (∀𝑥(𝜑𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓)))
87anbi2ci 731 . . 3 ((∀𝑥((𝜑𝜓) ∧ (𝜑 → ¬ 𝜓)) ∧ ∃𝑥𝜑) ↔ (∃𝑥𝜑 ∧ (∀𝑥(𝜑𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓))))
9 3anass 1040 . . 3 ((∃𝑥𝜑 ∧ ∀𝑥(𝜑𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓)) ↔ (∃𝑥𝜑 ∧ (∀𝑥(𝜑𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓))))
10 3anrot 1041 . . 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 312 1 ¬ (∀𝑥(𝜑𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 384  w3a 1036  wal 1478  wex 1701
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734
This theorem depends on definitions:  df-bi 197  df-an 386  df-3an 1038  df-ex 1702
This theorem is referenced by:  alsi-no-surprise  41845
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