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 ∨ ∀𝑥∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))

Proof of Theorem axbnd

Step	Hyp	Ref	Expression
1		nfnae 2306	. . . . . 6 ⊢ Ⅎ𝑥 ¬ ∀𝑧 𝑧 = 𝑥
2		nfnae 2306	. . . . . 6 ⊢ Ⅎ𝑥 ¬ ∀𝑧 𝑧 = 𝑦
3	1, 2	nfan 1816	. . . . 5 ⊢ Ⅎ𝑥(¬ ∀𝑧 𝑧 = 𝑥 ∧ ¬ ∀𝑧 𝑧 = 𝑦)
4		nfnae 2306	. . . . . . 7 ⊢ Ⅎ𝑧 ¬ ∀𝑧 𝑧 = 𝑥
5		nfnae 2306	. . . . . . 7 ⊢ Ⅎ𝑧 ¬ ∀𝑧 𝑧 = 𝑦
6	4, 5	nfan 1816	. . . . . 6 ⊢ Ⅎ𝑧(¬ ∀𝑧 𝑧 = 𝑥 ∧ ¬ ∀𝑧 𝑧 = 𝑦)
7		axc9 2290	. . . . . . 7 ⊢ (¬ ∀𝑧 𝑧 = 𝑥 → (¬ ∀𝑧 𝑧 = 𝑦 → (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
8	7	imp 444	. . . . . 6 ⊢ ((¬ ∀𝑧 𝑧 = 𝑥 ∧ ¬ ∀𝑧 𝑧 = 𝑦) → (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
9	6, 8	alrimi 2069	. . . . 5 ⊢ ((¬ ∀𝑧 𝑧 = 𝑥 ∧ ¬ ∀𝑧 𝑧 = 𝑦) → ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
10	3, 9	alrimi 2069	. . . 4 ⊢ ((¬ ∀𝑧 𝑧 = 𝑥 ∧ ¬ ∀𝑧 𝑧 = 𝑦) → ∀𝑥∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
11	10	ex 449	. . 3 ⊢ (¬ ∀𝑧 𝑧 = 𝑥 → (¬ ∀𝑧 𝑧 = 𝑦 → ∀𝑥∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
12	11	orrd 392	. 2 ⊢ (¬ ∀𝑧 𝑧 = 𝑥 → (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑥∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
13	12	orri 390	1 ⊢ (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑥∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))

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)