Let x0 of type ι → ι → ο be given.
Let x1 of type ι → ι → ο be given.
Let x2 of type ι → ι → ο be given.
Let x3 of type ι be given.
Let x4 of type ι be given.
Let x5 of type ι be given.
Let x6 of type ι be given.
Let x7 of type ι be given.
Let x8 of type ι → ι be given.
Let x9 of type ι → ι be given.
Let x10 of type ι → ι → ι be given.
Let x11 of type ι → ι → ι be given.
Let x12 of type ι → ι → ι be given.
Let x13 of type ι → ο be given.
Let x14 of type ι → ο be given.
Let x15 of type ι → ο be given.
Let x16 of type ι → ο be given.
Let x17 of type ι → ο be given.
Let x18 of type ι → ο be given.
Let x19 of type ι → ο be given.
Let x20 of type ι → ο be given.
Let x21 of type ι → ο be given.
Let x22 of type ι → ο be given.
Let x23 of type ι → ο be given.
Assume H0:
∀ x24 . iff (x19 x24) (and (x13 x24) (not (x14 x24))).
Assume H1:
∀ x24 . iff (x23 x24) (and (x17 x24) (not (x18 x24))).
Assume H2: ∀ x24 x25 . x1 x24 x25 ⟶ x1 x25 x24 ⟶ x24 = x25.
Assume H3:
∀ x24 . iff (x0 x24 x4) (x24 = x3).
Assume H4:
∀ x24 x25 x26 . iff (x0 x26 (x12 x24 x25)) (and (x0 x26 x24) (not (x0 x26 x25))).
Assume H5:
∀ x24 . x0 x24 x6 ⟶ or (or (x24 = x3) (x24 = x4)) (x24 = x5).
Assume H6: x0 x3 x7.
Assume H7: ∀ x24 . x0 x3 (x8 x24).
Assume H8: ∀ x24 x25 x26 . x10 x24 (x10 x25 x26) = x10 (x10 x24 x25) x26.
Assume H9: ∀ x24 x25 . x0 x24 x25 ⟶ x13 (x8 x25).
Assume H10:
∀ x24 x25 . not (x0 x25 x24) ⟶ x13 x24 ⟶ x14 (x10 x24 (x9 x25)).
Assume H11: ∀ x24 x25 . x0 x25 x24 ⟶ x1 (x9 x25) x24.
Assume H12: ∀ x24 x25 . x0 x25 x24 ⟶ x21 x24 ⟶ x20 (x12 x24 (x9 x25)).
Assume H13:
∀ x24 x25 x26 . x1 x26 (x10 x24 x25) ⟶ ∀ x27 . x0 x27 x24 ⟶ not (x0 x27 x26) ⟶ x15 x26 ⟶ or (x14 x24) (x14 x25).
Assume H14:
not (x0 x5 x3).
Assume H15:
not (x0 x6 x6).
Assume H16:
∀ x24 . x1 x24 x5 ⟶ not (x0 x3 x24) ⟶ x0 x4 x24 ⟶ x24 = x9 x4.
Assume H17: x0 x4 (x12 (x8 x5) (x8 (x9 x5))).
Assume H18:
not (x3 = x12 x6 (x9 x4)).
Assume H19: x0 (x9 x4) (x12 (x8 x5) (x9 x5)).
Assume H20:
not (x4 = x12 x6 (x9 x4)).
Assume H21:
not (x0 x4 (x12 x6 (x9 x4))).
Assume H22:
not (x1 (x9 x5) x5).
Assume H23:
not (x9 x4 = x12 (x8 x5) x5).
Assume H24: ∀ x24 . x0 x24 (x9 (x9 x4)) ⟶ x24 = x9 x4.