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 H2:
∀ x6 . x6 ∈ SNoS_ (SNoLev x0) ⟶ ∀ x7 . SNo x7 ⟶ x2 x6 x7 = x3 x6 x7.
Assume H3:
∀ x6 . x6 ∈ SNoS_ (SNoLev x1) ⟶ x2 x0 x6 = x3 x0 x6.
Apply SNoR_E with
x1,
x5,
add_SNo (x2 x4 x1) (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))) leaving 3 subgoals.
The subproof is completed by applying H1.
The subproof is completed by applying H4.
Apply H2 with
x4,
x1,
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))) leaving 3 subgoals.
The subproof is completed by applying H5.
The subproof is completed by applying H1.
Apply H2 with
x4,
x5,
λ x6 x7 . add_SNo (x3 x4 x1) (add_SNo (x2 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 H5.
The subproof is completed by applying H6.
Apply SNoR_SNoS_ with
x1,
x5.
The subproof is completed by applying H4.
Apply H3 with
x5,
λ x6 x7 . add_SNo (x3 x4 x1) (add_SNo x7 (minus_SNo (x3 x4 x5))) = add_SNo (x3 x4 x1) (add_SNo (x3 x0 x5) (minus_SNo (x3 x4 x5))) leaving 2 subgoals.
The subproof is completed by applying L9.
Let x6 of type ι → ι → ο be given.
The subproof is completed by applying H10.