Let x0 of type ι → ι → ο be given.
Assume H0: x0 0 4.
Assume H1: x0 4 5.
Assume H2: ∀ x1 x2 . x0 x1 x2 ⟶ x0 x2 x1.
Apply xm with
x0 0 5,
or (∃ x1 . and (x1 ⊆ 6) (and (equip 3 x1) (∀ x2 . x2 ∈ x1 ⟶ ∀ x3 . x3 ∈ x1 ⟶ (x2 = x3 ⟶ ∀ x4 : ο . x4) ⟶ x0 x2 x3))) (∃ x1 . and (x1 ⊆ 6) (and (equip 3 x1) (∀ x2 . x2 ∈ x1 ⟶ ∀ x3 . x3 ∈ x1 ⟶ (x2 = x3 ⟶ ∀ x4 : ο . x4) ⟶ not (x0 x2 x3)))) leaving 2 subgoals.
Assume H5: x0 0 5.
Apply orIL with
∃ x1 . and (x1 ⊆ 6) (and (equip 3 x1) (∀ x2 . x2 ∈ x1 ⟶ ∀ x3 . x3 ∈ x1 ⟶ (x2 = x3 ⟶ ∀ x4 : ο . x4) ⟶ x0 x2 x3)),
∃ x1 . and (x1 ⊆ 6) (and (equip 3 x1) (∀ x2 . x2 ∈ x1 ⟶ ∀ x3 . x3 ∈ x1 ⟶ (x2 = x3 ⟶ ∀ x4 : ο . x4) ⟶ not (x0 x2 x3))).
Apply L4 with
0 leaving 3 subgoals.
The subproof is completed by applying In_0_4.
The subproof is completed by applying H0.
The subproof is completed by applying H5.
Apply xm with
x0 1 4,
or (∃ x1 . and (x1 ⊆ 6) (and (equip 3 x1) (∀ x2 . x2 ∈ x1 ⟶ ∀ x3 . x3 ∈ x1 ⟶ (x2 = x3 ⟶ ∀ x4 : ο . x4) ⟶ x0 x2 x3))) (∃ x1 . and (x1 ⊆ 6) (and (equip 3 x1) (∀ x2 . x2 ∈ x1 ⟶ ∀ x3 . x3 ∈ x1 ⟶ (x2 = x3 ⟶ ∀ x4 : ο . x4) ⟶ not (x0 x2 x3)))) leaving 2 subgoals.
Assume H6: x0 1 4.
Apply xm with
x0 1 5,
or (∃ x1 . and (x1 ⊆ 6) (and (equip 3 x1) (∀ x2 . x2 ∈ x1 ⟶ ∀ x3 . x3 ∈ x1 ⟶ (x2 = x3 ⟶ ∀ x4 : ο . x4) ⟶ x0 x2 x3))) (∃ x1 . and (x1 ⊆ 6) (and (equip 3 x1) (∀ x2 . x2 ∈ x1 ⟶ ∀ x3 . x3 ∈ x1 ⟶ (x2 = x3 ⟶ ∀ x4 : ο . x4) ⟶ not (x0 x2 x3)))) leaving 2 subgoals.
Assume H7: x0 1 5.
Apply orIL with
∃ x1 . and (x1 ⊆ 6) (and (equip 3 x1) (∀ x2 . ... ⟶ ∀ x3 . x3 ∈ x1 ⟶ (x2 = x3 ⟶ ∀ x4 : ο . x4) ⟶ x0 x2 x3)),
....