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Theorem axbnd 2767
 Description: Axiom of Bundling (intuitionistic logic axiom ax-bnd). In classical logic, this and axi12 2765 are fairly straightforward consequences of axc9 2355. But in intuitionistic logic, it is not easy to add the extra ∀𝑥 to axi12 2765 and so we treat the two as separate axioms. (Contributed by Jim Kingdon, 22-Mar-2018.) (Proof shortened by Wolf Lammen, 24-Apr-2023.)
Assertion
Ref Expression
axbnd (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑥𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))

Proof of Theorem axbnd
StepHypRef Expression
1 nfae 2412 . . . . 5 𝑥𝑧 𝑧 = 𝑥
2 nfae 2412 . . . . 5 𝑥𝑧 𝑧 = 𝑦
31, 2nfor 1886 . . . 4 𝑥(∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦)
4319.32 2200 . . 3 (∀𝑥((∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦) ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)) ↔ ((∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦) ∨ ∀𝑥𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
5 orass 916 . . 3 (((∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦) ∨ ∀𝑥𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)) ↔ (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑥𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))))
64, 5bitri 276 . 2 (∀𝑥((∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦) ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)) ↔ (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑥𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))))
7 axi12 2765 . . 3 (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
8 orass 916 . . 3 (((∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦) ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)) ↔ (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))))
97, 8mpbir 232 . 2 ((∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦) ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
106, 9mpgbi 1780 1 (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑥𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ∨ wo 842  ∀wal 1520 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1777  ax-4 1791  ax-5 1888  ax-6 1947  ax-7 1992  ax-10 2112  ax-11 2126  ax-12 2141  ax-13 2344 This theorem depends on definitions:  df-bi 208  df-an 397  df-or 843  df-tru 1525  df-ex 1762  df-nf 1766 This theorem is referenced by: (None)
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