Let x0 of type ι be given.
Let x1 of type ι → ι → ο be given.
Assume H0: ∀ x2 . x2 ∈ x0 ⟶ ∀ x3 . x3 ∈ x0 ⟶ x1 x2 x3 ⟶ x1 x3 x2.
Let x2 of type ι be given.
Assume H1: x2 ∈ x0.
Let x3 of type ι be given.
Assume H2: x3 ∈ x0.
Let x4 of type ι be given.
Assume H3: x4 ∈ x0.
Let x5 of type ι be given.
Assume H4: x5 ∈ x0.
Let x6 of type ι be given.
Assume H5: x6 ∈ x0.
Assume H6:
87c36.. x1 x2 x3 x4 x5 x6.
Let x7 of type ο be given.
Assume H7:
2f869.. x1 x2 x4 x3 x6 ⟶ (x2 = x5 ⟶ ∀ x8 : ο . x8) ⟶ (x4 = x5 ⟶ ∀ x8 : ο . x8) ⟶ (x3 = x5 ⟶ ∀ x8 : ο . x8) ⟶ (x6 = x5 ⟶ ∀ x8 : ο . x8) ⟶ not (x1 x2 x5) ⟶ x1 x4 x5 ⟶ not (x1 x3 x5) ⟶ x1 x6 x5 ⟶ x7.
Apply H6 with
x7.
Apply H9 with
(x2 = x6 ⟶ ∀ x8 : ο . x8) ⟶ (x3 = x6 ⟶ ∀ x8 : ο . x8) ⟶ (x4 = x6 ⟶ ∀ x8 : ο . x8) ⟶ (x5 = x6 ⟶ ∀ x8 : ο . x8) ⟶ not (x1 x2 x6) ⟶ x1 x3 x6 ⟶ not (x1 x4 x6) ⟶ x1 x5 x6 ⟶ x7.
Assume H10: x2 = x3 ⟶ ∀ x8 : ο . x8.
Assume H11: x2 = x4 ⟶ ∀ x8 : ο . x8.
Assume H12: x3 = x4 ⟶ ∀ x8 : ο . x8.
Assume H13: x2 = x5 ⟶ ∀ x8 : ο . x8.
Assume H14: x3 = x5 ⟶ ∀ x8 : ο . x8.
Assume H15: x4 = x5 ⟶ ∀ x8 : ο . x8.
Assume H16:
not (x1 x2 x3).
Assume H17:
not (x1 x2 x4).
Assume H18:
not (x1 x3 x4).
Assume H19:
not (x1 x2 x5).
Assume H20:
not (x1 x3 x5).
Assume H21: x1 x4 x5.
Assume H22: x2 = x6 ⟶ ∀ x8 : ο . x8.
Assume H23: x3 = x6 ⟶ ∀ x8 : ο . x8.
Assume H24: x4 = x6 ⟶ ∀ x8 : ο . x8.
Assume H25: x5 = x6 ⟶ ∀ x8 : ο . x8.
Assume H26:
not (x1 x2 x6).
Assume H27: x1 x3 x6.
Assume H28:
not (x1 x4 x6).
Assume H29: x1 x5 x6.
Let x8 of type ο be given.
Assume H30:
... ⟶ ... ⟶ ... ⟶ ... ⟶ ... ⟶ ... ⟶ not (x1 ... ...) ⟶ not (x1 x2 x3) ⟶ not (x1 x4 x3) ⟶ not (x1 x2 x6) ⟶ not (x1 x4 x6) ⟶ x1 x3 x6 ⟶ x8.
Apply H7 leaving 9 subgoals.
The subproof is completed by applying L30.
The subproof is completed by applying H13.
The subproof is completed by applying H15.
The subproof is completed by applying H14.
Apply neq_i_sym with
x5,
x6.
The subproof is completed by applying H25.
The subproof is completed by applying H19.
The subproof is completed by applying H21.
The subproof is completed by applying H20.
Apply H0 with
x5,
x6 leaving 3 subgoals.
The subproof is completed by applying H4.
The subproof is completed by applying H5.
The subproof is completed by applying H29.