Metamath Proof Explorer

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