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: x0 x6 x7.
Assume H1:
∀ x24 x25 x26 . x0 x26 x24 ⟶ not (x0 x26 x25) ⟶ x0 x26 (x12 x24 x25).
Assume H2:
∀ x24 x25 . not (x0 x25 (x9 x24)) ⟶ not (x25 = x24).
Assume H3:
∀ x24 . not (x13 (x9 x24)).
Assume H4:
∀ x24 x25 x26 . x1 x26 (x10 x24 x25) ⟶ ∀ x27 . not (x0 x27 x26) ⟶ x1 x26 (x10 (x12 x24 (x9 x27)) x25).
Assume H5:
∀ x24 x25 x26 . x1 x26 (x10 x24 x25) ⟶ ∀ x27 . x0 x27 x24 ⟶ not (x0 x27 x26) ⟶ x15 x26 ⟶ x15 (x10 (x12 x24 (x9 x27)) x25).
Assume H6:
not (x0 x3 (x9 x5)).
Assume H7: x1 x4 (x12 x6 (x9 x4)).
Assume H8:
not (x0 x5 (x8 (x9 x4))).
Assume H9:
not (x1 x5 (x9 x5)).
Assume H10:
not (x3 = x10 (x9 x4) (x9 (x9 x4))).
Assume H11:
not (x1 (x12 (x8 x5) x5) x5).
Assume H12:
∀ x24 . x1 x24 (x8 (x9 x4)) ⟶ or (or (or (x24 = x3) (x24 = x4)) (x24 = x9 (x9 x4))) (x24 = x8 (x9 x4)).
Assume H13:
not (x0 (x8 x5) x6).
Assume H14:
not (x9 x5 = x6).
Assume H15: x0 (x9 (x9 x4)) (x12 (x8 (x8 (x9 x4))) (x9 (x8 (x9 x4)))).
Assume H16:
not (x0 (x9 x4) (x9 (x10 (x9 x4) (x9 (x9 x4))))).
Assume H17: x0 (x8 (x9 x4)) (x10 x6 (x9 (x8 (x9 x4)))).
Assume H18:
not (x0 (x12 x6 (x9 x4)) (x9 (x9 x5))).
Assume H19:
not (x0 x6 (x8 (x9 x5))).
Assume H20: x0 x3 (x10 (x9 x4) (x8 (x9 x5))).
Assume H21:
not (x0 (x9 x4) x7).
Assume H22: x0 (x9 x5) (x10 (x8 x5) (x9 (x9 x5))).
Assume H23:
not (x0 x5 (x10 x4 (x9 x6))).
Assume H24:
not (x0 (x9 x4) (x10 ... ...)).