∀n:nat, Q n == ∀n:nat, P (Z.of_nat n) ===> ∀x:Z, x <= 0 -> P x /\ || || (Q O) ∧ (∀n:nat, Q n -> Q (S n)) <=== (P 0) ∧ (∀x:Z, P x -> P (Z.succ x)) <=== (Z.of_nat (S n) = Z.succ (Z.of_nat n)) <=== Z_of_nat_completeThen the diagram will be closed and the theorem proved.