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Proofgold Proof

pf
Apply add_nat_SR with u26, 6, λ x0 x1 . x1 = u33 leaving 2 subgoals.
The subproof is completed by applying nat_6.
Apply add_nat_SR with u26, 5, λ x0 x1 . ordsucc x1 = u33 leaving 2 subgoals.
The subproof is completed by applying nat_5.
Apply add_nat_SR with u26, 4, λ x0 x1 . ordsucc (ordsucc x1) = u33 leaving 2 subgoals.
The subproof is completed by applying nat_4.
Apply add_nat_SR with u26, 3, λ x0 x1 . ordsucc (ordsucc (ordsucc x1)) = u33 leaving 2 subgoals.
The subproof is completed by applying nat_3.
Apply add_nat_SR with u26, 2, λ x0 x1 . ordsucc (ordsucc (ordsucc (ordsucc x1))) = u33 leaving 2 subgoals.
The subproof is completed by applying nat_2.
Apply add_nat_SR with u26, 1, λ x0 x1 . ordsucc (ordsucc (ordsucc (ordsucc (ordsucc x1)))) = u33 leaving 2 subgoals.
The subproof is completed by applying nat_1.
Apply add_nat_SR with u26, 0, λ x0 x1 . ordsucc (ordsucc (ordsucc (ordsucc (ordsucc (ordsucc x1))))) = u33 leaving 2 subgoals.
The subproof is completed by applying nat_0.
Apply add_nat_0R with u26, λ x0 x1 . ordsucc (ordsucc (ordsucc (ordsucc (ordsucc (ordsucc (ordsucc x1)))))) = u33.
Let x0 of type ιιο be given.
Assume H0: x0 (ordsucc (ordsucc (ordsucc (ordsucc (ordsucc (ordsucc (ordsucc u26))))))) u33.
The subproof is completed by applying H0.