Let x0 of type ι be given.

Let x1 of type ι → ι → ι be given.

Assume H0: ∀ x2 . prim1 x2 x0 ⟶ ∀ x3 . prim1 x3 x0 ⟶ prim1 (x1 x2 x3) x0.

Let x2 of type ι → ι → ι be given.

Assume H1: ∀ x3 . prim1 x3 x0 ⟶ ∀ x4 . prim1 x4 x0 ⟶ prim1 (x2 x3 x4) x0.

Let x3 of type ι → ι → ι be given.

Assume H2: ∀ x4 . prim1 x4 x0 ⟶ ∀ x5 . prim1 x5 x0 ⟶ prim1 (x3 x4 x5) x0.

Let x4 of type ι → ι be given.

Assume H3: ∀ x5 . prim1 x5 x0 ⟶ prim1 (x4 x5) x0.

Let x5 of type ι → ο be given.

Assume H4: ∀ x6 . ∀ x7 : ι → ι → ι . (∀ x8 . prim1 x8 x6 ⟶ ∀ x9 . prim1 x9 x6 ⟶ prim1 (x7 x8 x9) x6) ⟶ ∀ x8 : ι → ι → ι . (∀ x9 . prim1 x9 x6 ⟶ ∀ x10 . prim1 x10 x6 ⟶ prim1 (x8 x9 x10) x6) ⟶ ∀ x9 : ι → ι → ι . (∀ x10 . prim1 x10 x6 ⟶ ∀ x11 . prim1 x11 x6 ⟶ prim1 (x9 x10 x11) x6) ⟶ ∀ x10 : ι → ι . (∀ x11 . prim1 x11 x6 ⟶ prim1 (x10 x11) x6) ⟶ x5 (39199.. x6 x7 x8 x9 x10).

Apply H4 with x0, x1, x2, x3, x4 leaving 4 subgoals.

The subproof is completed by applying H0.

The subproof is completed by applying H1.

The subproof is completed by applying H2.

The subproof is completed by applying H3.

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Proofgold Proof