Let x0 of type ι → ο be given.
Let x1 of type ι → ι → ι be given.
Assume H0: ∀ x2 x3 . x0 x2 ⟶ x0 x3 ⟶ x0 (x1 x2 x3).
Assume H1: ∀ x2 x3 x4 . x0 x2 ⟶ x0 x3 ⟶ x0 x4 ⟶ x1 x2 (x1 x3 x4) = x1 x3 (x1 x2 x4).
Assume H2: ∀ x2 x3 . x0 x2 ⟶ x0 x3 ⟶ x1 x2 x3 = x1 x3 x2.
Let x2 of type ι 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.
Assume H3: x0 x2.
Assume H4: x0 x3.
Assume H5: x0 x4.
Assume H6: x0 x5.
Assume H7: x0 x6.
Apply H2 with
x5,
x6,
λ x7 x8 . x1 x2 (x1 x3 (x1 x4 x8)) = x1 x3 (x1 x2 (x1 x6 (x1 x4 x5))) leaving 3 subgoals.
The subproof is completed by applying H6.
The subproof is completed by applying H7.
Let x7 of type ι → ι → ο be given.
Apply unknownprop_309fec984ac6cc8b8adec797fb0a2d3f2693d312f2b022fd754734c930fe81be with
x0,
x1,
x3,
x2,
x6,
x4,
x5,
λ x8 x9 . x7 x9 x8 leaving 7 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H1.
The subproof is completed by applying H4.
The subproof is completed by applying H3.
The subproof is completed by applying H7.
The subproof is completed by applying H5.
The subproof is completed by applying H6.