Theorem axi12 2700

Description: Axiom of Quantifier Introduction (intuitionistic logic axiom ax-i12). In classical logic, this is mostly a restatement of axc9 2380 (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 2370. (Contributed by Jim Kingdon, 31-Dec-2017.) Avoid ax-11 2153. (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 2147	. . . . 5 ⊢ Ⅎ𝑧∀𝑧 𝑧 = 𝑥
2		nfa1 2147	. . . . 5 ⊢ Ⅎ𝑧∀𝑧 𝑧 = 𝑦
3	1, 2	nfor 1906	. . . 4 ⊢ Ⅎ𝑧(∀𝑧 𝑧 = 𝑥 ∨ ∀𝑧 𝑧 = 𝑦)
4	3	19.32 2225	. . 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 2380	. . . . . 6 ⊢ (¬ ∀𝑧 𝑧 = 𝑥 → (¬ ∀𝑧 𝑧 = 𝑦 → (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
6	5	orrd 860	. . . . 5 ⊢ (¬ ∀𝑧 𝑧 = 𝑥 → (∀𝑧 𝑧 = 𝑦 ∨ (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
7	6	orri 859	. . . 4 ⊢ (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
8		orass 919	. . . 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 230	. . 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 919	. 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 229	1 ⊢ (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))

Syntax hints:  ¬ wn 3   → wi 4   ∨ wo 844  ∀wal 1538

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 1912  ax-6 1970  ax-7 2010  ax-10 2136  ax-12 2170  ax-13 2370

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-tru 1543  df-ex 1781  df-nf 1785

This theorem is referenced by:  axbnd  2701