Let x0 of type ι be given.
Let x1 of type ι be given.
Let x2 of type ι → ι → ο be given.
Assume H0: ∀ x3 . x3 ∈ x1 ⟶ ∀ x4 . x4 ∈ x1 ⟶ x2 x3 x4 ⟶ x2 x4 x3.
Let x3 of type ι be given.
Assume H3: x3 ∈ x1.
Let x4 of type ι be given.
Assume H5: x4 ∈ x0.
Let x5 of type ι be given.
Assume H6: x5 ∈ x0.
Let x6 of type ι be given.
Assume H7: x6 ∈ x0.
Let x7 of type ι be given.
Assume H8: x7 ∈ x0.
Let x8 of type ι be given.
Assume H9: x8 ∈ x0.
Let x9 of type ι be given.
Assume H10: x9 ∈ x0.
Let x10 of type ι be given.
Assume H11: x10 ∈ x0.
Let x11 of type ι be given.
Assume H12: x11 ∈ x0.
Let x12 of type ι be given.
Assume H13: x12 ∈ x0.
Let x13 of type ι be given.
Assume H14: x13 ∈ x0.
Let x14 of type ι be given.
Assume H15: x14 ∈ x0.
Apply setminusE with
x1,
Sing x3,
x4,
0c718.. x2 x4 x5 x6 x7 x8 x9 x10 x11 x12 x13 x14 ⟶ ∀ x15 : ο . x15 leaving 2 subgoals.
Apply H4 with
x4.
The subproof is completed by applying H5.
Assume H16: x4 ∈ x1.
Apply setminusE with
x1,
Sing x3,
x5,
0c718.. x2 x4 x5 x6 x7 x8 x9 x10 x11 x12 x13 x14 ⟶ ∀ x15 : ο . x15 leaving 2 subgoals.
Apply H4 with
x5.
The subproof is completed by applying H6.
Assume H18: x5 ∈ x1.
Apply setminusE with
x1,
Sing x3,
x6,
0c718.. x2 x4 x5 x6 x7 x8 x9 x10 x11 x12 x13 x14 ⟶ ∀ x15 : ο . x15 leaving 2 subgoals.
Apply H4 with
x6.
The subproof is completed by applying H7.
Assume H20: x6 ∈ x1.
Apply setminusE with
x1,
Sing x3,
x7,
0c718.. x2 x4 x5 x6 x7 x8 x9 x10 x11 x12 x13 x14 ⟶ ∀ x15 : ο . x15 leaving 2 subgoals.
Apply H4 with
x7.
The subproof is completed by applying H8.
Assume H22: x7 ∈ x1.
Apply setminusE with
x1,
Sing x3,
x8,
0c718.. x2 x4 x5 x6 x7 x8 x9 x10 x11 x12 x13 x14 ⟶ ∀ x15 : ο . x15 leaving 2 subgoals.
Apply H4 with
x8.
The subproof is completed by applying H9.
Assume H24: x8 ∈ x1.
Apply setminusE with
x1,
Sing x3,
x9,
0c718.. x2 x4 x5 x6 x7 x8 x9 x10 x11 x12 x13 x14 ⟶ ∀ x15 : ο . x15 leaving 2 subgoals.
Apply H4 with
x9.
The subproof is completed by applying H10.
Assume H26: x9 ∈ x1.
Apply setminusE with
x1,
Sing x3,
x10,
0c718.. x2 x4 x5 x6 x7 x8 x9 x10 x11 x12 x13 x14 ⟶ ∀ x15 : ο . x15 leaving 2 subgoals.
Apply H4 with
x10.
The subproof is completed by applying H11.
Assume H28: x10 ∈ ....