Let x0 of type ι be given.
Let x1 of type ι be given.
Let x2 of type ι be given.
Let x3 of type ι be given.
Apply nat_ind with
λ x4 . ∀ x5 . equip x4 x5 ⟶ equip (add_nat x0 x4) (setsum x2 x5) leaving 2 subgoals.
Let x4 of type ι be given.
Apply add_nat_0R with
x0,
λ x5 x6 . equip x6 (setsum x2 x4).
Apply unknownprop_b1ab55fb3adc1f5608875c7150ec0901fc5d8b2cbdf1e5df6d670947db1f7326 with
x4,
λ x5 x6 . equip x0 (setsum x2 x6) leaving 2 subgoals.
The subproof is completed by applying H4.
Apply equip_tra with
x0,
x2,
setsum x2 0 leaving 2 subgoals.
The subproof is completed by applying H2.
Let x5 of type ο be given.
Assume H5:
∀ x6 : ι → ι . bij x2 (setsum x2 0) x6 ⟶ x5.
Apply H5 with
Inj0.
Apply bijI with
x2,
setsum x2 0,
Inj0 leaving 3 subgoals.
Let x6 of type ι be given.
Assume H6: x6 ∈ x2.
Apply Inj0_setsum with
x2,
0,
x6.
The subproof is completed by applying H6.
Let x6 of type ι be given.
Assume H6: x6 ∈ x2.
Let x7 of type ι be given.
Assume H7: x7 ∈ x2.
The subproof is completed by applying Inj0_inj with x6, x7.
Let x6 of type ι be given.
Assume H6:
x6 ∈ setsum x2 0.
Apply setsum_Inj_inv with
x2,
0,
x6,
∃ x7 . and (x7 ∈ x2) (Inj0 x7 = x6) leaving 3 subgoals.
The subproof is completed by applying H6.
Assume H7:
∃ x7 . and (x7 ∈ x2) (x6 = Inj0 x7).
Apply H7 with
∃ x7 . and (x7 ∈ x2) (Inj0 x7 = x6).
Let x7 of type ι be given.
Assume H8:
(λ x8 . and (x8 ∈ x2) (x6 = Inj0 x8)) x7.
Apply H8 with
∃ x8 . and (x8 ∈ x2) (Inj0 x8 = x6).
Assume H9: x7 ∈ x2.
Assume H10:
x6 = Inj0 x7.
Let x8 of type ο be given.
Assume H11:
∀ x9 . and (x9 ∈ x2) (Inj0 x9 = x6) ⟶ x8.
Apply H11 with
x7.
Apply andI with
x7 ∈ x2,
Inj0 x7 = x6 leaving 2 subgoals.
The subproof is completed by applying H9.
Let x9 of type ι → ι → ο be given.
The subproof is completed by applying H10 with λ x10 x11 . x9 x11 x10.
Assume H7:
∃ x7 . and (x7 ∈ 0) (x6 = Inj1 x7).
Apply H7 with
∃ x7 . and (x7 ∈ x2) (Inj0 x7 = x6).
Let x7 of type ι be given.
Assume H8:
(λ x8 . and (x8 ∈ 0) (x6 = Inj1 x8)) x7.
Apply H8 with
∃ x8 . and (x8 ∈ x2) (Inj0 x8 = x6).
Assume H9: x7 ∈ 0.
Apply FalseE with
x6 = Inj1 x7 ⟶ ∃ x8 . and (x8 ∈ x2) (Inj0 x8 = x6).
Apply EmptyE with
x7.
The subproof is completed by applying H9.
Let x4 of type ι be given.
Let x5 of type ι be given.
Assume H6: ....
Apply L4 with
x1,
x3 leaving 2 subgoals.
The subproof is completed by applying H1.
The subproof is completed by applying H3.