Theorem axbnd 2791

Description: Axiom of Bundling (intuitionistic logic axiom ax-bnd). In classical logic, this and axi12 2789 are fairly straightforward consequences of axc9 2396. But in intuitionistic logic, it is not easy to add the extra ∀𝑥 to axi12 2789 and so we treat the two as separate axioms. Usage of this theorem is discouraged because it depends on ax-13 2386. (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 2451	. . . . 5 ⊢ Ⅎ𝑥∀𝑧 𝑧 = 𝑥
2		nfae 2451	. . . . 5 ⊢ Ⅎ𝑥∀𝑧 𝑧 = 𝑦
3	1, 2	nfor 1901	. . . 4 ⊢ Ⅎ𝑥(∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦)
4	3	19.32 2230	. . 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 918	. . 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 277	. 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 2789	. . 3 ⊢ (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
8		orass 918	. . 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 233	. 2 ⊢ ((∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦) ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
10	6, 9	mpgbi 1795	1 ⊢ (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑥∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))

Syntax hints:   → wi 4   ∨ wo 843  ∀wal 1531

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-10 2141  ax-11 2156  ax-12 2172  ax-13 2386

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1536  df-ex 1777  df-nf 1781

This theorem is referenced by: (None)