∀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_complete
Then the diagram will be closed and the theorem proved.