Theorem axi12 2707

Description: Axiom of Quantifier Introduction (intuitionistic logic axiom ax-i12). In classical logic, this is mostly a restatement of axc9 2387 (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 2377. (Contributed by Jim Kingdon, 31-Dec-2017.) Avoid ax-11 2163. (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 2157	. . . . 5 ⊢ Ⅎ𝑧∀𝑧 𝑧 = 𝑥
2		nfa1 2157	. . . . 5 ⊢ Ⅎ𝑧∀𝑧 𝑧 = 𝑦
3	1, 2	nfor 1906	. . . 4 ⊢ Ⅎ𝑧(∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦)
4	3	19.32 2241	. . 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 2387	. . . . . 6 ⊢ (¬ ∀𝑧 𝑧 = 𝑥 → (¬ ∀𝑧 𝑧 = 𝑦 → (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
6	5	orrd 864	. . . . 5 ⊢ (¬ ∀𝑧 𝑧 = 𝑥 → (∀𝑧 𝑧 = 𝑦 ∨ (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
7	6	orri 863	. . . 4 ⊢ (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
8		orass 922	. . . 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 231	. . 3 ⊢ ((∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦) ∨ (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
10	4, 9	mpgbi 1800	. 2 ⊢ ((∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦) ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
11		orass 922	. 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 230	1 ⊢ (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))

Syntax hints:  ¬ wn 3   → wi 4   ∨ wo 848  ∀wal 1540

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-10 2147  ax-12 2185  ax-13 2377

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-ex 1782  df-nf 1786

This theorem is referenced by:  axbnd  2708