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.
Let x7 of type ι be given.
Assume H6: x7 ∈ x0.
Let x8 of type ι be given.
Assume H7: x8 ∈ x0.
Let x9 of type ι be given.
Assume H8: x9 ∈ x0.
Assume H9:
49663.. x1 x2 x3 x4 x5 x6 x7 x8 x9.
Let x10 of type ο be given.
Assume H10:
2319a.. x1 x2 x3 x4 x5 x7 x6 x8 ⟶ (x2 = x9 ⟶ ∀ x11 : ο . x11) ⟶ (x3 = x9 ⟶ ∀ x11 : ο . x11) ⟶ (x4 = x9 ⟶ ∀ x11 : ο . x11) ⟶ (x5 = x9 ⟶ ∀ x11 : ο . x11) ⟶ (x7 = x9 ⟶ ∀ x11 : ο . x11) ⟶ (x6 = x9 ⟶ ∀ x11 : ο . x11) ⟶ (x8 = x9 ⟶ ∀ x11 : ο . x11) ⟶ not (x1 x2 x9) ⟶ not (x1 x3 x9) ⟶ not (x1 x4 x9) ⟶ x1 x5 x9 ⟶ not (x1 x7 x9) ⟶ not (x1 x6 x9) ⟶ not (x1 x8 x9) ⟶ x10.
Apply H9 with
x10.
Assume H11:
2319a.. x1 x2 x3 x4 x5 x6 x7 x8.
Assume H12: x2 = x9 ⟶ ∀ x11 : ο . x11.
Assume H13: x3 = x9 ⟶ ∀ x11 : ο . x11.
Assume H14: x4 = x9 ⟶ ∀ x11 : ο . x11.
Assume H15: x5 = x9 ⟶ ∀ x11 : ο . x11.
Assume H16: x6 = x9 ⟶ ∀ x11 : ο . x11.
Assume H17: x7 = x9 ⟶ ∀ x11 : ο . x11.
Assume H18: x8 = x9 ⟶ ∀ x11 : ο . x11.
Assume H19:
not (x1 x2 x9).
Assume H20:
not (x1 x3 x9).
Assume H21:
not (x1 x4 x9).
Assume H22: x1 x5 x9.
Assume H23:
not (x1 x6 x9).
Assume H24:
not (x1 x7 x9).
Assume H25:
not (x1 x8 x9).
Apply H10 leaving 15 subgoals.
Apply unknownprop_e7c7302282f287c6fd29ba25f46d95230ea05b595d6a3b990d182d985d5a8cbf with
x0,
x1,
x2,
x3,
x4,
x5,
x6,
x7,
x8 leaving 9 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H1.
The subproof is completed by applying H2.
The subproof is completed by applying H3.
The subproof is completed by applying H4.
The subproof is completed by applying H5.
The subproof is completed by applying H6.
The subproof is completed by applying H7.
The subproof is completed by applying H11.
The subproof is completed by applying H12.
The subproof is completed by applying H13.
The subproof is completed by applying H14.
The subproof is completed by applying H15.
The subproof is completed by applying H17.
The subproof is completed by applying H16.
The subproof is completed by applying H18.
The subproof is completed by applying H19.
The subproof is completed by applying H20.
The subproof is completed by applying H21.
The subproof is completed by applying H22.
The subproof is completed by applying H24.
The subproof is completed by applying H23.
The subproof is completed by applying H25.