Theorem axbnd 2702

Description: Axiom of Bundling (intuitionistic logic axiom ax-bnd). In classical logic, this and axi12 2701 are fairly straightforward consequences of axc9 2382. But in intuitionistic logic, it is not easy to add the extra ∀𝑥 to axi12 2701 and so we treat the two as separate axioms. Usage of this theorem is discouraged because it depends on ax-13 2372. (Contributed by Jim Kingdon, 22-Mar-2018.) (Proof shortened by Wolf Lammen, 24-Apr-2023.) (New usage is discouraged.)

Assertion

Ref	Expression
axbnd	⊢ (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑥∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))

Proof of Theorem axbnd

Step	Hyp	Ref	Expression
1		nfae 2433	. . . . 5 ⊢ Ⅎ𝑥∀𝑧 𝑧 = 𝑥
2		nfae 2433	. . . . 5 ⊢ Ⅎ𝑥∀𝑧 𝑧 = 𝑦
3	1, 2	nfor 1905	. . . 4 ⊢ Ⅎ𝑥(∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦)
4	3	19.32 2236	. . 3 ⊢ (∀𝑥((∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦) ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)) ↔ ((∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦) ∨ ∀𝑥∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
5		orass 921	. . 3 ⊢ (((∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦) ∨ ∀𝑥∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)) ↔ (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑥∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))))
6	4, 5	bitri 275	. 2 ⊢ (∀𝑥((∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦) ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)) ↔ (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑥∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))))
7		axi12 2701	. . 3 ⊢ (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
8		orass 921	. . 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))))
9	7, 8	mpbir 231	. 2 ⊢ ((∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦) ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
10	6, 9	mpgbi 1799	1 ⊢ (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑥∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))

Syntax hints:   → wi 4   ∨ wo 847  ∀wal 1539

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-10 2144  ax-11 2160  ax-12 2180  ax-13 2372

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1544  df-ex 1781  df-nf 1785

This theorem is referenced by: (None)