Let x0 of type ι be given.
Apply xm with
∃ x1 . and (x1 ∈ x0) (x0 = ordsucc x1),
or (∀ x1 . x1 ∈ x0 ⟶ ordsucc x1 ∈ x0) (∃ x1 . and (x1 ∈ x0) (x0 = ordsucc x1)) leaving 2 subgoals.
Assume H1:
∃ x1 . and (x1 ∈ x0) (x0 = ordsucc x1).
Apply orIR with
∀ x1 . x1 ∈ x0 ⟶ ordsucc x1 ∈ x0,
∃ x1 . and (x1 ∈ x0) (x0 = ordsucc x1).
The subproof is completed by applying H1.
Apply orIL with
∀ x1 . x1 ∈ x0 ⟶ ordsucc x1 ∈ x0,
∃ x1 . and (x1 ∈ x0) (x0 = ordsucc x1).
Let x1 of type ι be given.
Assume H2: x1 ∈ x0.
Apply ordinal_ordsucc_In_eq with
x0,
x1,
ordsucc x1 ∈ x0 leaving 4 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H2.
The subproof is completed by applying H3.
Apply FalseE with
ordsucc x1 ∈ x0.
Apply H1.
Let x2 of type ο be given.
Assume H4:
∀ x3 . and (x3 ∈ x0) (x0 = ordsucc x3) ⟶ x2.
Apply H4 with
x1.
Apply andI with
x1 ∈ x0,
x0 = ordsucc x1 leaving 2 subgoals.
The subproof is completed by applying H2.
The subproof is completed by applying H3.