Let x0 of type ι → (ι → ((ι → ο) → ο) → ο) → ((ι → ο) → ο) → ο be given.
Assume H0:
∀ x1 . ∀ x2 x3 : ι → ((ι → ο) → ο) → ο . (∀ x4 . In x4 x1 ⟶ x2 x4 = x3 x4) ⟶ x0 x1 x2 = x0 x1 x3.
Apply unknownprop_acac0f89c78f08b97a9fe27ba4af5f929f74e43a9a77a0beb38d70975279c8b8 with
λ x1 . ∀ x2 x3 : ((ι → ο) → ο) → ο . 31b02.. x0 x1 x2 ⟶ 31b02.. x0 x1 x3 ⟶ x2 = x3.
Let x1 of type ι be given.
Assume H1:
∀ x2 . In x2 x1 ⟶ ∀ x3 x4 : ((ι → ο) → ο) → ο . 31b02.. x0 x2 x3 ⟶ 31b02.. x0 x2 x4 ⟶ x3 = x4.
Let x2 of type ((ι → ο) → ο) → ο be given.
Let x3 of type ((ι → ο) → ο) → ο be given.
Claim L4:
∃ x4 : ι → ((ι → ο) → ο) → ο . and (∀ x5 . In x5 x1 ⟶ 31b02.. x0 x5 (x4 x5)) (x2 = x0 x1 x4)Apply unknownprop_509d4e548f5be5c7cae0cd77af56983da0a4fd9c38f78b9d02b09d350f5a6017 with
x0,
x1,
x2.
The subproof is completed by applying H2.
Claim L5:
∃ x4 : ι → ((ι → ο) → ο) → ο . and (∀ x5 . In x5 x1 ⟶ 31b02.. x0 x5 (x4 x5)) (x3 = x0 x1 x4)Apply unknownprop_509d4e548f5be5c7cae0cd77af56983da0a4fd9c38f78b9d02b09d350f5a6017 with
x0,
x1,
x3.
The subproof is completed by applying H3.
Apply L4 with
x2 = x3.
Let x4 of type ι → ((ι → ο) → ο) → ο be given.
Assume H6:
(λ x5 : ι → ((ι → ο) → ο) → ο . and (∀ x6 . In x6 x1 ⟶ 31b02.. x0 x6 (x5 x6)) (x2 = x0 x1 x5)) x4.
Apply andE with
∀ x5 . In x5 x1 ⟶ 31b02.. x0 x5 (x4 x5),
x2 = x0 x1 x4,
x2 = x3 leaving 2 subgoals.
The subproof is completed by applying H6.
Assume H7:
∀ x5 . In x5 x1 ⟶ 31b02.. x0 x5 (x4 x5).
Assume H8: x2 = x0 x1 x4.
Apply L5 with
x2 = x3.
Let x5 of type ι → ((ι → ο) → ο) → ο be given.
Assume H9:
(λ x6 : ι → ((ι → ο) → ο) → ο . and (∀ x7 . In x7 x1 ⟶ 31b02.. x0 x7 (x6 x7)) (x3 = x0 x1 x6)) x5.
Apply andE with
∀ x6 . In x6 x1 ⟶ 31b02.. x0 x6 (x5 x6),
x3 = x0 x1 x5,
x2 = x3 leaving 2 subgoals.
The subproof is completed by applying H9.
Assume H10:
∀ x6 . In x6 x1 ⟶ 31b02.. x0 x6 (x5 x6).
Assume H11: x3 = x0 x1 x5.
Apply H8 with
λ x6 x7 : ((ι → ο) → ο) → ο . x7 = x3.
Apply H11 with
λ x6 x7 : ((ι → ο) → ο) → ο . 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 H12.
Apply H7 with
x6.
The subproof is completed by applying H12.
Apply H10 with
x6.
The subproof is completed by applying H12.