Let x0 of type ι be given.
Let x1 of type ι be given.
Assume H2:
∀ x2 . x2 ∈ x1 ⟶ ordinal x2.
Apply H1 with
∃ x2 : ι → ι . and (inj x0 x1 x2) (∀ x3 . x3 ∈ x0 ⟶ ∀ x4 . x4 ∈ x3 ⟶ x2 x4 ∈ x2 x3).
Let x2 of type ι → ι be given.
Apply H3 with
∃ x3 : ι → ι . and (inj x0 x1 x3) (∀ x4 . x4 ∈ x0 ⟶ ∀ x5 . x5 ∈ x4 ⟶ x3 x5 ∈ x3 x4).
Assume H4: ∀ x3 . x3 ∈ x0 ⟶ x2 x3 ∈ x1.
Assume H5: ∀ x3 . x3 ∈ x0 ⟶ ∀ x4 . x4 ∈ x0 ⟶ x2 x3 = x2 x4 ⟶ x3 = x4.
Claim L6: {...|x3 ∈ x0} ⊆ ...
Apply unknownprop_d5c45d9008e943c9d649bff721cdcc4661b243f2e11e35ea7a42c8fe41f77f58 with
x0,
{x2 x3|x3 ∈ x0},
∃ x3 : ι → ι . and (inj x0 x1 x3) (∀ x4 . x4 ∈ x0 ⟶ ∀ x5 . x5 ∈ x4 ⟶ x3 x5 ∈ x3 x4) leaving 4 subgoals.
The subproof is completed by applying H0.
Apply unknownprop_6f924010899e62355200d41f1cef23d6373bef28ff540d0bdb872dcb6e86d39f with
x0,
x2.
The subproof is completed by applying H5.
Let x3 of type ι be given.
Assume H7: x3 ∈ {x2 x4|x4 ∈ x0}.
Apply H2 with
x3.
Apply L6 with
x3.
The subproof is completed by applying H7.
Let x3 of type ι → ι be given.
Assume H7:
(λ x4 : ι → ι . and (inj x0 (prim5 x0 x2) x4) (∀ x5 . x5 ∈ x0 ⟶ ∀ x6 . x6 ∈ x5 ⟶ x4 x6 ∈ x4 x5)) x3.
Apply H7 with
∃ x4 : ι → ι . and (inj x0 x1 x4) (∀ x5 . x5 ∈ x0 ⟶ ∀ x6 . x6 ∈ x5 ⟶ x4 x6 ∈ x4 x5).
Assume H8:
inj x0 {x2 x4|x4 ∈ x0} x3.
Assume H9: ∀ x4 . x4 ∈ x0 ⟶ ∀ x5 . x5 ∈ x4 ⟶ x3 x5 ∈ x3 x4.
Let x4 of type ο be given.
Assume H10:
∀ x5 : ι → ι . and (inj x0 x1 x5) (∀ x6 . x6 ∈ x0 ⟶ ∀ x7 . x7 ∈ x6 ⟶ x5 x7 ∈ x5 x6) ⟶ x4.
Apply H10 with
x3.
Apply andI with
inj x0 x1 x3,
∀ x5 . x5 ∈ x0 ⟶ ∀ x6 . x6 ∈ x5 ⟶ x3 x6 ∈ x3 x5 leaving 2 subgoals.
Apply H8 with
inj x0 x1 x3.
Assume H11: ∀ x5 . x5 ∈ x0 ⟶ x3 x5 ∈ {x2 x6|x6 ∈ x0}.
Apply andI with
∀ x5 . x5 ∈ x0 ⟶ x3 x5 ∈ x1,
∀ x5 . x5 ∈ x0 ⟶ ∀ x6 . x6 ∈ x0 ⟶ x3 x5 = x3 x6 ⟶ x5 = x6.
Let x5 of type ι be given.
Assume H12: x5 ∈ x0.
Apply L6 with
x3 x5.
Apply H11 with
x5.
The subproof is completed by applying H12.
The subproof is completed by applying H9.