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Theorem axbnd 2588
Description: Axiom of Bundling (intuitionistic logic axiom ax-bnd). In classical logic, this and axi12 2587 are fairly straightforward consequences of axc9 2289. But in intuitionistic logic, it is not easy to add the extra 𝑥 to axi12 2587 and so we treat the two as separate axioms. (Contributed by Jim Kingdon, 22-Mar-2018.)
Assertion
Ref Expression
axbnd (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑥𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))

Proof of Theorem axbnd
StepHypRef Expression
1 nfnae 2305 . . . . . 6 𝑥 ¬ ∀𝑧 𝑧 = 𝑥
2 nfnae 2305 . . . . . 6 𝑥 ¬ ∀𝑧 𝑧 = 𝑦
31, 2nfan 1815 . . . . 5 𝑥(¬ ∀𝑧 𝑧 = 𝑥 ∧ ¬ ∀𝑧 𝑧 = 𝑦)
4 nfnae 2305 . . . . . . 7 𝑧 ¬ ∀𝑧 𝑧 = 𝑥
5 nfnae 2305 . . . . . . 7 𝑧 ¬ ∀𝑧 𝑧 = 𝑦
64, 5nfan 1815 . . . . . 6 𝑧(¬ ∀𝑧 𝑧 = 𝑥 ∧ ¬ ∀𝑧 𝑧 = 𝑦)
7 axc9 2289 . . . . . . 7 (¬ ∀𝑧 𝑧 = 𝑥 → (¬ ∀𝑧 𝑧 = 𝑦 → (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
87imp 443 . . . . . 6 ((¬ ∀𝑧 𝑧 = 𝑥 ∧ ¬ ∀𝑧 𝑧 = 𝑦) → (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
96, 8alrimi 2068 . . . . 5 ((¬ ∀𝑧 𝑧 = 𝑥 ∧ ¬ ∀𝑧 𝑧 = 𝑦) → ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
103, 9alrimi 2068 . . . 4 ((¬ ∀𝑧 𝑧 = 𝑥 ∧ ¬ ∀𝑧 𝑧 = 𝑦) → ∀𝑥𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
1110ex 448 . . 3 (¬ ∀𝑧 𝑧 = 𝑥 → (¬ ∀𝑧 𝑧 = 𝑦 → ∀𝑥𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
1211orrd 391 . 2 (¬ ∀𝑧 𝑧 = 𝑥 → (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑥𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
1312orri 389 1 (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑥𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wo 381  wa 382  wal 1472
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1712  ax-4 1727  ax-5 1826  ax-6 1874  ax-7 1921  ax-10 2005  ax-11 2020  ax-12 2033  ax-13 2233
This theorem depends on definitions:  df-bi 195  df-or 383  df-an 384  df-tru 1477  df-ex 1695  df-nf 1700
This theorem is referenced by: (None)
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