Theorem axbnd

Description: Axiom of Bundling (intuitionistic logic axiom ax-bnd). In classical logic, this and axi12 are fairly straightforward consequences of axc9 . But in intuitionistic logic, it is not easy to add the extra A. x to axi12 and so we treat the two as separate axioms. Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by Jim Kingdon, 22-Mar-2018) (Proof shortened by Wolf Lammen, 24-Apr-2023) (New usage is discouraged.)

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

Proof

Step	Hyp	Ref	Expression
1		nfae	⊢ Ⅎ 𝑥 ∀ 𝑧 𝑧 = 𝑥
2		nfae	⊢ Ⅎ 𝑥 ∀ 𝑧 𝑧 = 𝑦
3	1 2	nfor	⊢ Ⅎ 𝑥 ( ∀ 𝑧 𝑧 = 𝑥 ∨ ∀ 𝑧 𝑧 = 𝑦 )
4	3	19.32	⊢ ( ∀ 𝑥 ( ( ∀ 𝑧 𝑧 = 𝑥 ∨ ∀ 𝑧 𝑧 = 𝑦 ) ∨ ∀ 𝑧 ( 𝑥 = 𝑦 → ∀ 𝑧 𝑥 = 𝑦 ) ) ↔ ( ( ∀ 𝑧 𝑧 = 𝑥 ∨ ∀ 𝑧 𝑧 = 𝑦 ) ∨ ∀ 𝑥 ∀ 𝑧 ( 𝑥 = 𝑦 → ∀ 𝑧 𝑥 = 𝑦 ) ) )
5		orass	⊢ ( ( ( ∀ 𝑧 𝑧 = 𝑥 ∨ ∀ 𝑧 𝑧 = 𝑦 ) ∨ ∀ 𝑥 ∀ 𝑧 ( 𝑥 = 𝑦 → ∀ 𝑧 𝑥 = 𝑦 ) ) ↔ ( ∀ 𝑧 𝑧 = 𝑥 ∨ ( ∀ 𝑧 𝑧 = 𝑦 ∨ ∀ 𝑥 ∀ 𝑧 ( 𝑥 = 𝑦 → ∀ 𝑧 𝑥 = 𝑦 ) ) ) )
6	4 5	bitri	⊢ ( ∀ 𝑥 ( ( ∀ 𝑧 𝑧 = 𝑥 ∨ ∀ 𝑧 𝑧 = 𝑦 ) ∨ ∀ 𝑧 ( 𝑥 = 𝑦 → ∀ 𝑧 𝑥 = 𝑦 ) ) ↔ ( ∀ 𝑧 𝑧 = 𝑥 ∨ ( ∀ 𝑧 𝑧 = 𝑦 ∨ ∀ 𝑥 ∀ 𝑧 ( 𝑥 = 𝑦 → ∀ 𝑧 𝑥 = 𝑦 ) ) ) )
7		axi12	⊢ ( ∀ 𝑧 𝑧 = 𝑥 ∨ ( ∀ 𝑧 𝑧 = 𝑦 ∨ ∀ 𝑧 ( 𝑥 = 𝑦 → ∀ 𝑧 𝑥 = 𝑦 ) ) )
8		orass	⊢ ( ( ( ∀ 𝑧 𝑧 = 𝑥 ∨ ∀ 𝑧 𝑧 = 𝑦 ) ∨ ∀ 𝑧 ( 𝑥 = 𝑦 → ∀ 𝑧 𝑥 = 𝑦 ) ) ↔ ( ∀ 𝑧 𝑧 = 𝑥 ∨ ( ∀ 𝑧 𝑧 = 𝑦 ∨ ∀ 𝑧 ( 𝑥 = 𝑦 → ∀ 𝑧 𝑥 = 𝑦 ) ) ) )
9	7 8	mpbir	⊢ ( ( ∀ 𝑧 𝑧 = 𝑥 ∨ ∀ 𝑧 𝑧 = 𝑦 ) ∨ ∀ 𝑧 ( 𝑥 = 𝑦 → ∀ 𝑧 𝑥 = 𝑦 ) )
10	6 9	mpgbi	⊢ ( ∀ 𝑧 𝑧 = 𝑥 ∨ ( ∀ 𝑧 𝑧 = 𝑦 ∨ ∀ 𝑥 ∀ 𝑧 ( 𝑥 = 𝑦 → ∀ 𝑧 𝑥 = 𝑦 ) ) )