let n, m be Nat; ex k being Nat st
( (divSeq (m,n)) . k = 0 & (modSeq (m,n)) . k = 0 )
set f = modSeq (m,n);
consider k being Nat such that
A1:
(modSeq (m,n)) . k = 0
by Lm4;
take
k + 1
; ( (divSeq (m,n)) . (k + 1) = 0 & (modSeq (m,n)) . (k + 1) = 0 )
A2:
k + 0 < k + 1
by XREAL_1:6;
hence
(divSeq (m,n)) . (k + 1) = 0
by A1, Th18; (modSeq (m,n)) . (k + 1) = 0
thus
(modSeq (m,n)) . (k + 1) = 0
by A1, A2, Th14; verum