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.
Apply setminusE with
x1,
Sing x3,
x4,
f96b5.. x2 x4 x5 x6 x7 x8 x9 x10 x11 x12 ⟶ ∀ x13 : ο . (x2 x4 x3 ⟶ not (x2 x5 x3) ⟶ not (x2 x6 x3) ⟶ not (x2 x7 x3) ⟶ x2 x8 x3 ⟶ not (x2 x9 x3) ⟶ not (x2 x10 x3) ⟶ not (x2 x11 x3) ⟶ not (x2 x12 x3) ⟶ x13) ⟶ (x2 x4 x3 ⟶ not (x2 x5 x3) ⟶ not (x2 x6 x3) ⟶ not (x2 x7 x3) ⟶ x2 x8 x3 ⟶ not (x2 x9 x3) ⟶ not (x2 x10 x3) ⟶ not (x2 x11 x3) ⟶ x2 x12 x3 ⟶ x13) ⟶ x13 leaving 2 subgoals.
Apply H4 with
x4.
The subproof is completed by applying H5.
Assume H14: x4 ∈ x1.
Apply setminusE with
x1,
Sing x3,
x5,
f96b5.. x2 x4 x5 x6 x7 x8 x9 x10 x11 x12 ⟶ ∀ x13 : ο . (x2 x4 x3 ⟶ not (x2 x5 x3) ⟶ not (x2 x6 x3) ⟶ not (x2 x7 x3) ⟶ x2 x8 x3 ⟶ not (x2 x9 x3) ⟶ not (x2 x10 x3) ⟶ not (x2 x11 x3) ⟶ not (x2 x12 x3) ⟶ x13) ⟶ (x2 x4 x3 ⟶ not (x2 x5 x3) ⟶ not (x2 x6 x3) ⟶ not (x2 x7 x3) ⟶ x2 x8 x3 ⟶ not (x2 x9 x3) ⟶ not (x2 x10 x3) ⟶ not (x2 x11 x3) ⟶ x2 x12 x3 ⟶ x13) ⟶ x13 leaving 2 subgoals.
Apply H4 with
x5.
The subproof is completed by applying H6.