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Theorem ralnralall 4444
Description: A contradiction concerning restricted generalization for a nonempty set implies anything. (Contributed by Alexander van der Vekens, 4-Sep-2018.)
Assertion
Ref Expression
ralnralall (𝐴 ≠ ∅ → ((∀𝑥𝐴 𝜑 ∧ ∀𝑥𝐴 ¬ 𝜑) → 𝜓))
Distinct variable group:   𝑥,𝐴
Allowed substitution hints:   𝜑(𝑥)   𝜓(𝑥)

Proof of Theorem ralnralall
StepHypRef Expression
1 r19.26 3101 . 2 (∀𝑥𝐴 (𝜑 ∧ ¬ 𝜑) ↔ (∀𝑥𝐴 𝜑 ∧ ∀𝑥𝐴 ¬ 𝜑))
2 pm3.24 404 . . . . 5 ¬ (𝜑 ∧ ¬ 𝜑)
32bifal 1564 . . . 4 ((𝜑 ∧ ¬ 𝜑) ↔ ⊥)
43ralbii 3087 . . 3 (∀𝑥𝐴 (𝜑 ∧ ¬ 𝜑) ↔ ∀𝑥𝐴 ⊥)
5 r19.3rzv 4434 . . . 4 (𝐴 ≠ ∅ → (⊥ ↔ ∀𝑥𝐴 ⊥))
6 falim 1565 . . . 4 (⊥ → 𝜓)
75, 6biimtrrdi 256 . . 3 (𝐴 ≠ ∅ → (∀𝑥𝐴 ⊥ → 𝜓))
84, 7biimtrid 244 . 2 (𝐴 ≠ ∅ → (∀𝑥𝐴 (𝜑 ∧ ¬ 𝜑) → 𝜓))
91, 8biimtrrid 245 1 (𝐴 ≠ ∅ → ((∀𝑥𝐴 𝜑 ∧ ∀𝑥𝐴 ¬ 𝜑) → 𝜓))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 397  wfal 1560  wne 2936  wral 3055  c0 4264
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1975  ax-7 2016  ax-9 2131  ax-ext 2713
This theorem depends on definitions:  df-bi 209  df-an 398  df-tru 1551  df-fal 1561  df-ex 1788  df-sb 2075  df-clab 2720  df-cleq 2733  df-ne 2937  df-ral 3056  df-dif 3888  df-nul 4265
This theorem is referenced by: (None)
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