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Proofgold Proof

pf
Apply add_nat_SR with u19, 5, λ x0 x1 . x1 = u25 leaving 2 subgoals.
The subproof is completed by applying nat_5.
Apply add_nat_SR with u19, 4, λ x0 x1 . ordsucc x1 = u25 leaving 2 subgoals.
The subproof is completed by applying nat_4.
Apply add_nat_SR with u19, 3, λ x0 x1 . ordsucc (ordsucc x1) = u25 leaving 2 subgoals.
The subproof is completed by applying nat_3.
Apply add_nat_SR with u19, 2, λ x0 x1 . ordsucc (ordsucc (ordsucc x1)) = u25 leaving 2 subgoals.
The subproof is completed by applying nat_2.
Apply add_nat_SR with u19, 1, λ x0 x1 . ordsucc (ordsucc (ordsucc (ordsucc x1))) = u25 leaving 2 subgoals.
The subproof is completed by applying nat_1.
Apply add_nat_SR with u19, 0, λ x0 x1 . ordsucc (ordsucc (ordsucc (ordsucc (ordsucc x1)))) = u25 leaving 2 subgoals.
The subproof is completed by applying nat_0.
Apply add_nat_0R with u19, λ x0 x1 . ordsucc (ordsucc (ordsucc (ordsucc (ordsucc (ordsucc x1))))) = u25.
Let x0 of type ιιο be given.
Assume H0: x0 (ordsucc (ordsucc (ordsucc (ordsucc (ordsucc (ordsucc u19)))))) u25.
The subproof is completed by applying H0.