Check alpha_nat_divisibility_2.
Print Assumptions alpha_nat_divisibility_2.
Section congruence_1.
Variable (b1 b2 : nat) (Hb1 : 2 <= b1) (Hb2 : 2 <= b2)
(q : nat) (Hq : q <> 0) (Hb : nat2Zp Hq b1 = nat2Zp Hq b2).
Hint Resolve Zle_0_nat.

Theorem alpha_nat_congruence_0 b1 b2 q n :
2 <= b1
-> 2 <= b2
-> rem b1 q = rem b2 q
-> rem (alpha_nat b1 n) q = rem (alpha_nat b2 n) q.