Let x0 of type ι → ο be given.
Let x1 of type ι be given.
Apply unknownprop_24d3fab8c5f2be313e198261133fcc4de3913b4e3dd8f154e7592564c2e0ab41 with
λ x2 x3 : ι → ο . x3 x1 ⟶ x0 x1.
Assume H3:
(λ x2 . ∀ x3 : ι → ο . x3 (Inj0 0) ⟶ x3 (Inj0 (Power 0)) ⟶ (∀ x4 x5 . x3 x4 ⟶ x3 x5 ⟶ x3 (Inj1 (setsum x4 x5))) ⟶ x3 x2) x1.
Apply H3 with
λ x2 . and (combinator x2) (x0 x2) leaving 3 subgoals.
Apply unknownprop_389e2fb1855352fcc964ea44fe6723d7a1c2d512f04685300e3e97621725b977 with
combinator (Inj0 0),
x0 (Inj0 0) leaving 2 subgoals.
The subproof is completed by applying unknownprop_1be6ee89e47bf30db415892dfc0492321aa3b37926a377afe8fdc4c0df645b89.
The subproof is completed by applying H0.
Apply unknownprop_389e2fb1855352fcc964ea44fe6723d7a1c2d512f04685300e3e97621725b977 with
combinator (Inj0 (Power 0)),
x0 (Inj0 (Power 0)) leaving 2 subgoals.
The subproof is completed by applying unknownprop_2def6145ac5421f9112c2d9b9900cdb9fe5a639cc9f5e2053ff582b1b8cee0be.
The subproof is completed by applying H1.
Let x2 of type ι be given.
Let x3 of type ι be given.
Apply andE with
combinator x2,
x0 x2,
(λ x4 . and (combinator x4) (x0 x4)) (Inj1 (setsum x2 x3)) leaving 2 subgoals.
The subproof is completed by applying H4.
Assume H7: x0 x2.
Apply andE with
combinator x3,
x0 x3,
(λ x4 . and (combinator x4) (x0 x4)) (Inj1 (setsum x2 x3)) leaving 2 subgoals.
The subproof is completed by applying H5.
Assume H9: x0 x3.
Apply unknownprop_389e2fb1855352fcc964ea44fe6723d7a1c2d512f04685300e3e97621725b977 with
combinator ((λ x4 x5 . Inj1 (setsum x4 x5)) x2 x3),
x0 ((λ x4 x5 . Inj1 (setsum x4 x5)) x2 x3) leaving 2 subgoals.
Apply unknownprop_640ecaf2a8c271a844d2072fe17fde7f6008bced6f0262b37c69e34afaea64bc with
x2,
x3 leaving 2 subgoals.
The subproof is completed by applying H6.
The subproof is completed by applying H8.
Apply H2 with
x2,
x3 leaving 4 subgoals.
The subproof is completed by applying H6.
The subproof is completed by applying H7.
The subproof is completed by applying H8.
The subproof is completed by applying H9.
Apply unknownprop_896ccc9f209efa8a895211d65adb5a90348b419f100f6ab5e9762ce4d7fa9cc1 with
combinator x1,
x0 x1.
The subproof is completed by applying L4.