Let x0 of type ι be given.
Let x1 of type ι be given.
Let x2 of type ι be given.
Let x3 of type ι → ι → ο be given.
Assume H1: ∀ x4 x5 . x3 x4 x5 ⟶ x3 x5 x4.
Apply H0 with
λ x4 x5 . not (x3 x4 x5),
or (∃ x4 . and (x4 ⊆ x2) (and (atleastp x1 x4) (∀ x5 . x5 ∈ x4 ⟶ ∀ x6 . x6 ∈ x4 ⟶ (x5 = x6 ⟶ ∀ x7 : ο . x7) ⟶ x3 x5 x6))) (∃ x4 . and (x4 ⊆ x2) (and (atleastp x0 x4) (∀ x5 . x5 ∈ x4 ⟶ ∀ x6 . x6 ∈ x4 ⟶ (x5 = x6 ⟶ ∀ x7 : ο . x7) ⟶ not (x3 x5 x6)))) leaving 3 subgoals.
The subproof is completed by applying L2.
Assume H3:
∃ x4 . and (x4 ⊆ x2) (and (atleastp x0 x4) (∀ x5 . x5 ∈ x4 ⟶ ∀ x6 . x6 ∈ x4 ⟶ (x5 = x6 ⟶ ∀ x7 : ο . x7) ⟶ not (x3 x5 x6))).
Apply orIR with
∃ x4 . and (x4 ⊆ x2) (and (atleastp x1 x4) (∀ x5 . x5 ∈ x4 ⟶ ∀ x6 . x6 ∈ x4 ⟶ (x5 = x6 ⟶ ∀ x7 : ο . x7) ⟶ x3 x5 x6)),
∃ x4 . and (x4 ⊆ x2) (and (atleastp x0 x4) (∀ x5 . x5 ∈ x4 ⟶ ∀ x6 . x6 ∈ x4 ⟶ (x5 = x6 ⟶ ∀ x7 : ο . x7) ⟶ not (x3 x5 x6))).
The subproof is completed by applying H3.
Assume H3:
∃ x4 . and (x4 ⊆ x2) (and (atleastp x1 x4) (∀ x5 . x5 ∈ x4 ⟶ ∀ x6 . x6 ∈ x4 ⟶ (x5 = x6 ⟶ ∀ x7 : ο . x7) ⟶ not (not (x3 x5 x6)))).
Apply H3 with
or (∃ x4 . and (x4 ⊆ x2) (and (atleastp x1 x4) (∀ x5 . x5 ∈ x4 ⟶ ∀ x6 . x6 ∈ x4 ⟶ (x5 = x6 ⟶ ∀ x7 : ο . x7) ⟶ x3 x5 x6))) (∃ x4 . and (x4 ⊆ x2) (and (atleastp x0 x4) (∀ x5 . x5 ∈ x4 ⟶ ∀ x6 . x6 ∈ x4 ⟶ (x5 = x6 ⟶ ∀ x7 : ο . x7) ⟶ not (x3 x5 x6)))).
Let x4 of type ι be given.
Assume H4:
(λ x5 . and (x5 ⊆ x2) (and (atleastp x1 x5) (∀ x6 . x6 ∈ x5 ⟶ ∀ x7 . x7 ∈ x5 ⟶ (x6 = x7 ⟶ ∀ x8 : ο . x8) ⟶ not (not (x3 x6 x7))))) x4.
Apply H4 with
or (∃ x5 . and (x5 ⊆ x2) (and (atleastp x1 x5) (∀ x6 . ... ⟶ ∀ x7 . ...))) ....