Let x0 of type ι → ι → ο be given.
Assume H0: ∀ x1 x2 . x0 x1 x2 ⟶ x0 x2 x1.
Apply xm with
x0 4 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 H1: x0 4 5.
Apply unknownprop_6a985c2bc7f0f34dcb9cbf722df17cdf501e7087fca990e1fbbdb3baf2586ba2 with
x0 leaving 2 subgoals.
The subproof is completed by applying H1.
The subproof is completed by applying H0.
Apply unknownprop_6a985c2bc7f0f34dcb9cbf722df17cdf501e7087fca990e1fbbdb3baf2586ba2 with
λ x1 x2 . not (x0 x1 x2),
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 4 subgoals.
The subproof is completed by applying L2.
The subproof is completed by applying L3.
The subproof is completed by applying orIR 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
λ x1 x2 : ι → ι → ο . (∃ x3 . and (x3 ⊆ 6) (and (equip 3 x3) (∀ x4 . x4 ∈ x3 ⟶ ∀ x5 . x5 ∈ x3 ⟶ (x4 = x5 ⟶ ∀ x6 : ο . x6) ⟶ x2 x4 x5))) ⟶ or (∃ x3 . and (x3 ⊆ 6) (and (equip 3 x3) (∀ x4 . x4 ∈ x3 ⟶ ∀ x5 . x5 ∈ x3 ⟶ (x4 = x5 ⟶ ∀ x6 : ο . x6) ⟶ x0 x4 x5))) (∃ x3 . and (x3 ⊆ 6) (and (equip 3 x3) (∀ x4 . x4 ∈ x3 ⟶ ∀ x5 . x5 ∈ x3 ⟶ (x4 = x5 ⟶ ∀ x6 : ο . x6) ⟶ not (x0 x4 x5)))).
The subproof is completed by applying orIL with
∃ x1 . and (x1 ⊆ 6) (and (equip 3 x1) (∀ x2 . ...)),
....