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Apply nat_ind with λ x0 . ∀ x1 x2 : ι → ι . (∀ x3 . x3 ∈ ordsucc x0 ⟶ SNo (x1 x3)) ⟶ (∀ x3 . x3 ∈ ordsucc x0 ⟶ SNo (x2 x3)) ⟶ x1 (ordsucc x0) = x1 0 ⟶ (∀ x3 . x3 ∈ ordsucc x0 ⟶ SNoLe (add_SNo (x1 (ordsucc x3)) (minus_SNo (x1 x3))) (x2 x3)) ⟶ SNoLe 0 (finite_add_SNo (ordsucc x0) x2) leaving 2 subgoals.
Let x0 of type ι → ι be given.
Let x1 of type ι → ι be given.
Assume H0: ∀ x2 . x2 ∈ 1 ⟶ SNo (x0 x2).
Assume H1: ∀ x2 . x2 ∈ 1 ⟶ SNo (x1 x2).
Assume H2: x0 1 = x0 0.
Apply H2 with λ x2 x3 . add_SNo x3 (minus_SNo (x0 0)) = 0.
Apply add_SNo_minus_SNo_rinv with x0 0.
Apply H0 with 0.
The subproof is completed by applying In_0_1.
Apply unknownprop_69faeba9a75f2f8d58865148c9f0f7e35d3ac66d15111e3cc3404d6a2eed4dcd with x1, 0, λ x2 x3 . SNoLe 0 x3 leaving 2 subgoals.
The subproof is completed by applying nat_0.
Apply unknownprop_4376a6c44e07ee63cfd63de35739aa1967e60551544bcab6d1b8284e5b2ad2ba with x1, λ x2 x3 . SNoLe 0 (add_SNo x3 (x1 0)).
Apply add_SNo_0L with x1 0, λ x2 x3 . SNoLe 0 x3 leaving 2 subgoals.
Apply H1 with 0.
The subproof is completed by applying In_0_1.
Apply L4 with λ x2 x3 . SNoLe x2 (x1 0).
Apply H3 with 0.
The subproof is completed by applying In_0_1.
Let x0 of type ι be given.
Let x1 of type ι → ι be given.
Let x2 of type ι → ι be given.
Apply unknownprop_69faeba9a75f2f8d58865148c9f0f7e35d3ac66d15111e3cc3404d6a2eed4dcd with x2, ordsucc x0, λ x3 x4 . SNoLe 0 x4 leaving 2 subgoals.
Apply nat_ordsucc with x0.
The subproof is completed by applying H0.
Apply unknownprop_69faeba9a75f2f8d58865148c9f0f7e35d3ac66d15111e3cc3404d6a2eed4dcd with x2, x0, λ x3 x4 . SNoLe 0 (add_SNo x4 (x2 (ordsucc x0))) leaving 2 subgoals.
The subproof is completed by applying H0.
Apply add_SNo_assoc with finite_add_SNo x0 x2, x2 x0, x2 (ordsucc x0), λ x3 x4 . SNoLe 0 x3 leaving 4 subgoals.
Apply unknownprop_814c004451946f1c9f980b657692a7768a826c8b8434cb9bb69b9f9613c2c671 with x0, x2 leaving 2 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying L10.
The subproof is completed by applying L11.
The subproof is completed by applying L12.
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