Let x0 of type ι be given.

Let x1 of type ι be given.

Let x2 of type ι → ι → ι be given.

Let x3 of type ι → ι → ι be given.

Let x4 of type ι be given.

Let x5 of type ι be given.

Assume H0: ∀ x6 . x6 ∈ SNoS_ (SNoLev x0) ⟶ ∀ x7 . SNo x7 ⟶ x2 x6 x7 = x3 x6 x7.

Assume H1: ∀ x6 . x6 ∈ SNoS_ (SNoLev x1) ⟶ x2 x0 x6 = x3 x0 x6.

Assume H2: x4 ∈ SNoS_ (SNoLev x0).

Assume H3: SNo x5.

Assume H4: x2 x4 x1 = x3 x4 x1.

Assume H5: x2 x0 x5 = x3 x0 x5.

Apply H4 with λ x6 x7 . add_SNo x7 (add_SNo (x2 x0 x5) (minus_SNo (x2 x4 x5))) = add_SNo (x3 x4 x1) (add_SNo (x3 x0 x5) (minus_SNo (x3 x4 x5))).

Apply H5 with λ x6 x7 . add_SNo (x3 x4 x1) (add_SNo x7 (minus_SNo (x2 x4 x5))) = add_SNo (x3 x4 x1) (add_SNo (x3 x0 x5) (minus_SNo (x3 x4 x5))).

Apply H0 with x4, x5, λ x6 x7 . add_SNo (x3 x4 x1) (add_SNo (x3 x0 x5) (minus_SNo x7)) = add_SNo (x3 x4 x1) (add_SNo (x3 x0 x5) (minus_SNo (x3 x4 x5))) leaving 3 subgoals.

The subproof is completed by applying H2.

The subproof is completed by applying H3.

Let x6 of type ι → ι → ο be given.

Assume H6: x6 (add_SNo (x3 x4 x1) (add_SNo (x3 x0 x5) (minus_SNo (x3 x4 x5)))) (add_SNo (x3 x4 x1) (add_SNo (x3 x0 x5) (minus_SNo (x3 x4 x5)))).

The subproof is completed by applying H6.

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Proofgold Proof