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_Z_congr n : Z2Zp Hq (alpha_Z b1 n) = Z2Zp Hq (alpha_Z b2 n).
End congruence_1.