Let x0 of type ι → (ι → ι → ι → ι) → ι → ι → ι be given.
Assume H0:
∀ x1 . ∀ x2 x3 : ι → ι → ι → ι . (∀ x4 . In x4 x1 ⟶ x2 x4 = x3 x4) ⟶ x0 x1 x2 = x0 x1 x3.
Apply unknownprop_acac0f89c78f08b97a9fe27ba4af5f929f74e43a9a77a0beb38d70975279c8b8 with
λ x1 . ∀ x2 x3 : ι → ι → ι . ca584.. x0 x1 x2 ⟶ ca584.. x0 x1 x3 ⟶ x2 = x3.
Let x1 of type ι be given.
Assume H1:
∀ x2 . In x2 x1 ⟶ ∀ x3 x4 : ι → ι → ι . ca584.. x0 x2 x3 ⟶ ca584.. x0 x2 x4 ⟶ x3 = x4.
Let x2 of type ι → ι → ι be given.
Let x3 of type ι → ι → ι be given.
Claim L4:
∃ x4 : ι → ι → ι → ι . and (∀ x5 . In x5 x1 ⟶ ca584.. x0 x5 (x4 x5)) (x2 = x0 x1 x4)Apply unknownprop_158ac6211240830ed5fc7a82c1748019ed48f1067c60f0cbad0e47a403c8e2fb with
x0,
x1,
x2.
The subproof is completed by applying H2.
Claim L5:
∃ x4 : ι → ι → ι → ι . and (∀ x5 . In x5 x1 ⟶ ca584.. x0 x5 (x4 x5)) (x3 = x0 x1 x4)Apply unknownprop_158ac6211240830ed5fc7a82c1748019ed48f1067c60f0cbad0e47a403c8e2fb with
x0,
x1,
x3.
The subproof is completed by applying H3.
Apply unknownprop_b46ca74db86c4d63f341bbc962aa4cedea2bc3c0ecd5b9cc51d3353d4d2e115b with
λ x4 : ι → ι → ι → ι . ∀ x5 . In x5 x1 ⟶ ca584.. x0 x5 (x4 x5),
λ x4 : ι → ι → ι → ι . x2 = x0 x1 x4,
x2 = x3 leaving 2 subgoals.
The subproof is completed by applying L4.
Let x4 of type ι → ι → ι → ι be given.
Assume H6:
∀ x5 . In x5 x1 ⟶ ca584.. x0 x5 (x4 x5).
Assume H7: x2 = x0 x1 x4.
Apply unknownprop_b46ca74db86c4d63f341bbc962aa4cedea2bc3c0ecd5b9cc51d3353d4d2e115b with
λ x5 : ι → ι → ι → ι . ∀ x6 . In x6 x1 ⟶ ca584.. x0 x6 (x5 x6),
λ x5 : ι → ι → ι → ι . x3 = x0 x1 x5,
x2 = x3 leaving 2 subgoals.
The subproof is completed by applying L5.
Let x5 of type ι → ι → ι → ι be given.
Assume H8:
∀ x6 . In x6 x1 ⟶ ca584.. x0 x6 (x5 x6).
Assume H9: x3 = x0 x1 x5.
Apply H7 with
λ x6 x7 : ι → ι → ι . x7 = x3.
Apply H9 with
λ x6 x7 : ι → ι → ι . x0 x1 x4 = x7.
Apply H0 with
x1,
x4,
x5.
Let x6 of type ι be given.
Apply H1 with
x6,
x4 x6,
x5 x6 leaving 3 subgoals.
The subproof is completed by applying H10.
Apply H6 with
x6.
The subproof is completed by applying H10.
Apply H8 with
x6.
The subproof is completed by applying H10.