Let x0 of type ι → ι → ο be given.
Assume H0: ∀ x1 x2 . x0 x1 x2 ⟶ x0 x2 x1.
Assume H1:
not (∃ x1 . and (x1 ⊆ u9) (and (equip u3 x1) (∀ x2 . x2 ∈ x1 ⟶ ∀ x3 . x3 ∈ x1 ⟶ (x2 = x3 ⟶ ∀ x4 : ο . x4) ⟶ x0 x2 x3))).
Let x1 of type ι be given.
Let x2 of type ι be given.
Let x3 of type ι be given.
Assume H5: x0 x1 x2.
Assume H6: x0 x1 x3.
Assume H7: x0 x2 x3.
Let x4 of type ο be given.
Assume H8: x1 = x2 ⟶ x4.
Assume H9: x1 = x3 ⟶ x4.
Assume H10: x2 = x3 ⟶ x4.
Apply dneg with
x4.
Apply H1.
Let x5 of type ο be given.
Assume H13:
∀ x6 . and (x6 ⊆ u9) (and (equip u3 x6) (∀ x7 . x7 ∈ x6 ⟶ ∀ x8 . x8 ∈ x6 ⟶ (x7 = x8 ⟶ ∀ x9 : ο . x9) ⟶ x0 x7 x8)) ⟶ x5.
Apply H13 with
SetAdjoin (UPair x1 x2) x3.
Apply andI with
SetAdjoin (UPair x1 x2) x3 ⊆ u9,
and (equip u3 (SetAdjoin (UPair x1 x2) x3)) (∀ x6 . x6 ∈ SetAdjoin (UPair x1 x2) x3 ⟶ ∀ x7 . x7 ∈ SetAdjoin (UPair x1 x2) x3 ⟶ (x6 = x7 ⟶ ∀ x8 : ο . x8) ⟶ x0 x6 x7) leaving 2 subgoals.
Let x6 of type ι be given.
Apply L12 with
x6,
λ x7 . x7 ∈ u9 leaving 4 subgoals.
The subproof is completed by applying H14.
The subproof is completed by applying H2.
The subproof is completed by applying H3.
The subproof is completed by applying H4.
Apply andI with
equip u3 (SetAdjoin (UPair x1 x2) x3),
∀ x6 . x6 ∈ SetAdjoin (UPair x1 x2) x3 ⟶ ∀ x7 . x7 ∈ SetAdjoin (UPair x1 x2) x3 ⟶ (x6 = x7 ⟶ ∀ x8 : ο . x8) ⟶ x0 x6 x7 leaving 2 subgoals.
Apply equip_sym with
SetAdjoin (UPair x1 x2) x3,
u3.
Apply unknownprop_01a9c78d2ff9508973a3397619af294eba02d9395696c331bc156cf4e0508f7d with
x1,
x2,
x3 leaving 3 subgoals.
Assume H14: x1 = x2.
Apply H11.
Apply H8.
The subproof is completed by applying H14.
Assume H14: x1 = x3.
Apply H11.
Apply H9.
The subproof is completed by applying H14.
Assume H14: x2 = x3.
Apply H11.
Apply H10.
The subproof is completed by applying H14.
Let x6 of type ι be given.
Apply L12 with
x6,
λ x7 . ∀ x8 . x8 ∈ SetAdjoin (UPair x1 x2) x3 ⟶ (x7 = x8 ⟶ ∀ x9 : ο . x9) ⟶ x0 x7 x8 leaving 4 subgoals.
The subproof is completed by applying H14.
Let x7 of type ι be given.
Apply L12 with
x7,
λ x8 . (x1 = x8 ⟶ ∀ x9 : ο . x9) ⟶ x0 x1 x8 leaving 4 subgoals.
The subproof is completed by applying H15.
Assume H16: x1 = x1 ⟶ ∀ x8 : ο . x8.
Apply FalseE with
x0 ... ....