Theorem axbnd 2601

Description: Axiom of Bundling (intuitionistic logic axiom ax-bnd). In classical logic, this and axi12 2600 are fairly straightforward consequences of axc9 2302. But in intuitionistic logic, it is not easy to add the extra ∀𝑥 to axi12 2600 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 ∨ ∀𝑥∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))

Proof of Theorem axbnd

Step	Hyp	Ref	Expression
1		nfnae 2318	. . . . . 6 ⊢ Ⅎ𝑥 ¬ ∀𝑧 𝑧 = 𝑥
2		nfnae 2318	. . . . . 6 ⊢ Ⅎ𝑥 ¬ ∀𝑧 𝑧 = 𝑦
3	1, 2	nfan 1828	. . . . 5 ⊢ Ⅎ𝑥(¬ ∀𝑧 𝑧 = 𝑥 ∧ ¬ ∀𝑧 𝑧 = 𝑦)
4		nfnae 2318	. . . . . . 7 ⊢ Ⅎ𝑧 ¬ ∀𝑧 𝑧 = 𝑥
5		nfnae 2318	. . . . . . 7 ⊢ Ⅎ𝑧 ¬ ∀𝑧 𝑧 = 𝑦
6	4, 5	nfan 1828	. . . . . 6 ⊢ Ⅎ𝑧(¬ ∀𝑧 𝑧 = 𝑥 ∧ ¬ ∀𝑧 𝑧 = 𝑦)
7		axc9 2302	. . . . . . 7 ⊢ (¬ ∀𝑧 𝑧 = 𝑥 → (¬ ∀𝑧 𝑧 = 𝑦 → (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
8	7	imp 445	. . . . . 6 ⊢ ((¬ ∀𝑧 𝑧 = 𝑥 ∧ ¬ ∀𝑧 𝑧 = 𝑦) → (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
9	6, 8	alrimi 2082	. . . . 5 ⊢ ((¬ ∀𝑧 𝑧 = 𝑥 ∧ ¬ ∀𝑧 𝑧 = 𝑦) → ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
10	3, 9	alrimi 2082	. . . 4 ⊢ ((¬ ∀𝑧 𝑧 = 𝑥 ∧ ¬ ∀𝑧 𝑧 = 𝑦) → ∀𝑥∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
11	10	ex 450	. . 3 ⊢ (¬ ∀𝑧 𝑧 = 𝑥 → (¬ ∀𝑧 𝑧 = 𝑦 → ∀𝑥∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
12	11	orrd 393	. 2 ⊢ (¬ ∀𝑧 𝑧 = 𝑥 → (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑥∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
13	12	orri 391	1 ⊢ (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑥∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))

Syntax hints:  ¬ wn 3   → wi 4   ∨ wo 383   ∧ wa 384  ∀wal 1481

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710

This theorem is referenced by: (None)