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.
Let x10 of type ι be given.
Assume H9: x10 ∈ x0.
Let x11 of type ι be given.
Assume H10: x11 ∈ x0.
Assume H11:
d4ea7.. x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 x11.
Let x12 of type ο be given.
Assume H12:
76a6c.. x1 x2 x3 x5 x4 x6 x7 x8 x9 x10 ⟶ (x2 = x11 ⟶ ∀ x13 : ο . x13) ⟶ (x3 = x11 ⟶ ∀ x13 : ο . x13) ⟶ (x5 = x11 ⟶ ∀ x13 : ο . x13) ⟶ (x4 = x11 ⟶ ∀ x13 : ο . x13) ⟶ (x6 = x11 ⟶ ∀ x13 : ο . x13) ⟶ (x7 = x11 ⟶ ∀ x13 : ο . x13) ⟶ (x8 = x11 ⟶ ∀ x13 : ο . x13) ⟶ (x9 = x11 ⟶ ∀ x13 : ο . x13) ⟶ (x10 = x11 ⟶ ∀ x13 : ο . x13) ⟶ x1 x2 x11 ⟶ not (x1 x3 x11) ⟶ not (x1 x5 x11) ⟶ not (x1 x4 x11) ⟶ x1 x6 x11 ⟶ x1 x7 x11 ⟶ not (x1 x8 x11) ⟶ x1 x9 x11 ⟶ not (x1 x10 x11) ⟶ x12.
Apply H11 with
x12.
Assume H13:
76a6c.. x1 x2 x3 x4 x5 x6 x7 x8 x9 x10.
Assume H14: x2 = x11 ⟶ ∀ x13 : ο . x13.
Assume H15: x3 = x11 ⟶ ∀ x13 : ο . x13.
Assume H16: x4 = x11 ⟶ ∀ x13 : ο . x13.
Assume H17: x5 = x11 ⟶ ∀ x13 : ο . x13.
Assume H18: x6 = x11 ⟶ ∀ x13 : ο . x13.
Assume H19: x7 = x11 ⟶ ∀ x13 : ο . x13.
Assume H20: x8 = x11 ⟶ ∀ x13 : ο . x13.
Assume H21: x9 = x11 ⟶ ∀ x13 : ο . x13.
Assume H22: x10 = x11 ⟶ ∀ x13 : ο . x13.
Assume H23: x1 x2 x11.
Assume H24:
not (x1 x3 x11).
Assume H25:
not (x1 x4 x11).
Assume H26:
not (x1 x5 x11).
Assume H27: x1 x6 x11.
Assume H28: x1 x7 x11.
Assume H29:
not (x1 x8 x11).
Assume H30: x1 x9 x11.
Assume H31:
not (x1 x10 x11).
Apply H12 leaving 19 subgoals.
Apply unknownprop_f9064b3053aaf724d572f0448a70632a5d208abd2eb4bd7402c31c75dd8007fd with
x0,
x1,
x2,
x3,
x4,
x5,
x6,
x7,
x8,
x9,
x10 leaving 11 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.