Let x0 of type ι → ο be given.
Let x1 of type ι → ι → ι → ο be given.
Let x2 of type ι → ι be given.
Let x3 of type ι → ι → ι → ι → ι → ι be given.
Let x4 of type ι → ο be given.
Let x5 of type ι → ι → ι → ο be given.
Let x6 of type ι → ι be given.
Let x7 of type ι → ι → ι → ι → ι → ι be given.
Let x8 of type ι → ι be given.
Let x9 of type ι → ι → ι → ι be given.
Let x10 of type ι → ι be given.
Let x11 of type ι → ι → ι → ι be given.
Let x12 of type ι → ι be given.
Let x13 of type ι → ι be given.
The subproof is completed by applying andI with ∀ x14 . x0 x14 ⟶ x7 (x8 x14) (x8 (x10 (x8 x14))) (x8 x14) (x13 (x8 x14)) (x9 x14 (x10 (x8 x14)) (x12 x14)) = x6 (x8 x14), ∀ x14 . x4 x14 ⟶ x3 (x10 x14) (x10 (x8 (x10 x14))) (x10 x14) (x11 (x8 (x10 x14)) x14 (x13 x14)) (x12 (x10 x14)) = x2 (x10 x14).