Let x0 of type ι be given.
Let x1 of type ι → ο be given.
Assume H0:
∀ x2 . x1 x2 ⟶ ∀ x3 . x3 ∈ x2 ⟶ nIn x0 x3.
Let x2 of type ι → ι be given.
Let x3 of type ι → ι be given.
Let x4 of type ι → ι → ι be given.
Let x5 of type ι → ι → ι be given.
Assume H1: x1 0.
Assume H2: ∀ x6 . x1 x6 ⟶ x1 (x3 x6).
Assume H3: x2 0 = 0.
Assume H4: x4 0 0 = 0.
Assume H5: ∀ x6 . x1 x6 ⟶ x5 0 x6 = 0.
Assume H6: ∀ x6 . x1 x6 ⟶ x5 x6 0 = 0.
Let x6 of type ι be given.
Apply CD_proj0_F with
x0,
x1,
0,
λ x7 x8 . pair_tag x0 (x4 (x5 x8 (CD_proj0 x0 x1 x6)) (x2 (x5 (x3 (CD_proj1 x0 x1 x6)) (CD_proj1 x0 x1 0)))) (x4 (x5 (CD_proj1 x0 x1 x6) x8) (x5 (CD_proj1 x0 x1 0) (x3 (CD_proj0 x0 x1 x6)))) = 0 leaving 4 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H1.
The subproof is completed by applying H1.
Apply CD_proj1_F with
x0,
x1,
0,
λ x7 x8 . pair_tag x0 (x4 (x5 0 (CD_proj0 x0 x1 x6)) (x2 (x5 (x3 (CD_proj1 x0 x1 x6)) x8))) (x4 (x5 (CD_proj1 x0 x1 x6) 0) (x5 x8 (x3 (CD_proj0 x0 x1 x6)))) = 0 leaving 4 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H1.
The subproof is completed by applying H1.
Apply H5 with
CD_proj0 x0 x1 x6,
λ x7 x8 . pair_tag x0 (x4 x8 (x2 (x5 (x3 (CD_proj1 x0 x1 x6)) 0))) (x4 (x5 (CD_proj1 x0 x1 x6) 0) (x5 0 (x3 (CD_proj0 x0 x1 x6)))) = 0 leaving 2 subgoals.
The subproof is completed by applying L8.
Apply H5 with
x3 (CD_proj0 x0 x1 x6),
λ x7 x8 . pair_tag x0 (x4 0 (x2 (x5 (x3 (CD_proj1 x0 x1 x6)) 0))) (x4 (x5 (CD_proj1 x0 x1 x6) 0) x8) = 0 leaving 2 subgoals.
The subproof is completed by applying L10.
Apply H6 with
x3 (CD_proj1 x0 x1 x6),
λ x7 x8 . pair_tag x0 (x4 0 (x2 x8)) (x4 (x5 (CD_proj1 x0 x1 x6) 0) 0) = 0 leaving 2 subgoals.
The subproof is completed by applying L11.
Apply H6 with
CD_proj1 x0 x1 x6,
λ x7 x8 . pair_tag x0 (x4 0 (x2 0)) (x4 x8 0) = 0 leaving 2 subgoals.
The subproof is completed by applying L9.
Apply H3 with
λ x7 x8 . pair_tag x0 (x4 0 x8) (x4 0 0) = 0.
Apply H4 with
λ x7 x8 . pair_tag x0 x8 x8 = 0.
Apply pair_tag_0 with
x0,
x1,
0.
The subproof is completed by applying H0.