let a, b, m, n be Nat; ( a < b & (modSeq m,n) . a = 0 implies (divSeq m,n) . b = 0 )
set fd = divSeq m,n;
set fm = modSeq m,n;
assume
a < b
; ( not (modSeq m,n) . a = 0 or (divSeq m,n) . b = 0 )
then A1:
a + 1 <= b
by NAT_1:13;
assume
(modSeq m,n) . a = 0
; (divSeq m,n) . b = 0
then
(divSeq m,n) . (a + 1) = 0
by Lm4;
hence
(divSeq m,n) . b = 0
by A1, Th17; verum