Theorem axi12 2791

Description: Axiom of Quantifier Introduction (intuitionistic logic axiom ax-i12). In classical logic, this is mostly a restatement of axc9 2400 (with one additional quantifier). But in intuitionistic logic, changing the negations and implications to disjunctions makes it stronger. Usage of this theorem is discouraged because it depends on ax-13 2390. (Contributed by Jim Kingdon, 31-Dec-2017.) Avoid ax-11 2161. (Revised by Wolf Lammen, 24-Apr-2023.) (New usage is discouraged.)

Assertion

Ref	Expression
axi12	⊢ (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))

Proof of Theorem axi12

Step	Hyp	Ref	Expression
1		nfa1 2155	. . . . 5 ⊢ Ⅎ𝑧∀𝑧 𝑧 = 𝑥
2		nfa1 2155	. . . . 5 ⊢ Ⅎ𝑧∀𝑧 𝑧 = 𝑦
3	1, 2	nfor 1905	. . . 4 ⊢ Ⅎ𝑧(∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦)
4	3	19.32 2235	. . 3 ⊢ (∀𝑧((∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦) ∨ (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)) ↔ ((∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦) ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
5		axc9 2400	. . . . . 6 ⊢ (¬ ∀𝑧 𝑧 = 𝑥 → (¬ ∀𝑧 𝑧 = 𝑦 → (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
6	5	orrd 859	. . . . 5 ⊢ (¬ ∀𝑧 𝑧 = 𝑥 → (∀𝑧 𝑧 = 𝑦 ∨ (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
7	6	orri 858	. . . 4 ⊢ (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
8		orass 918	. . . 4 ⊢ (((∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦) ∨ (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)) ↔ (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))))
9	7, 8	mpbir 233	. . 3 ⊢ ((∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦) ∨ (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
10	4, 9	mpgbi 1799	. 2 ⊢ ((∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦) ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
11		orass 918	. 2 ⊢ (((∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦) ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)) ↔ (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))))
12	10, 11	mpbi 232	1 ⊢ (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))

Syntax hints:  ¬ wn 3   → wi 4   ∨ wo 843  ∀wal 1535

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 1970  ax-7 2015  ax-10 2145  ax-12 2177  ax-13 2390

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1540  df-ex 1781  df-nf 1785

This theorem is referenced by:  axbnd  2793