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Theorem axbnd 2589
 Description: Axiom of Bundling (intuitionistic logic axiom ax-bnd). In classical logic, this and axi12 2588 are fairly straightforward consequences of axc9 2290. But in intuitionistic logic, it is not easy to add the extra ∀𝑥 to axi12 2588 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 2306 . . . . . 6 𝑥 ¬ ∀𝑧 𝑧 = 𝑥
2 nfnae 2306 . . . . . 6 𝑥 ¬ ∀𝑧 𝑧 = 𝑦
31, 2nfan 1816 . . . . 5 𝑥(¬ ∀𝑧 𝑧 = 𝑥 ∧ ¬ ∀𝑧 𝑧 = 𝑦)
4 nfnae 2306 . . . . . . 7 𝑧 ¬ ∀𝑧 𝑧 = 𝑥
5 nfnae 2306 . . . . . . 7 𝑧 ¬ ∀𝑧 𝑧 = 𝑦
64, 5nfan 1816 . . . . . 6 𝑧(¬ ∀𝑧 𝑧 = 𝑥 ∧ ¬ ∀𝑧 𝑧 = 𝑦)
7 axc9 2290 . . . . . . 7 (¬ ∀𝑧 𝑧 = 𝑥 → (¬ ∀𝑧 𝑧 = 𝑦 → (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
87imp 444 . . . . . 6 ((¬ ∀𝑧 𝑧 = 𝑥 ∧ ¬ ∀𝑧 𝑧 = 𝑦) → (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
96, 8alrimi 2069 . . . . 5 ((¬ ∀𝑧 𝑧 = 𝑥 ∧ ¬ ∀𝑧 𝑧 = 𝑦) → ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
103, 9alrimi 2069 . . . 4 ((¬ ∀𝑧 𝑧 = 𝑥 ∧ ¬ ∀𝑧 𝑧 = 𝑦) → ∀𝑥𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
1110ex 449 . . 3 (¬ ∀𝑧 𝑧 = 𝑥 → (¬ ∀𝑧 𝑧 = 𝑦 → ∀𝑥𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
1211orrd 392 . 2 (¬ ∀𝑧 𝑧 = 𝑥 → (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑥𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
1312orri 390 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 382   ∧ wa 383  ∀wal 1473 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234 This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701 This theorem is referenced by: (None)
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