thn m. let a,b,c E N | a|b, b|c, a|c. prf. assume a,b,c E N | a|b, b|c. by 'divides' Z+ d,e such that ad=b and be=c. by substitution, c=be - _ m Xx X X X // cc xx c=(ad)e. by assoc. prop. of mult. for Z+, (ad)e= a(de). by trans. prop. c=a(de). by closure of mult Z+, (de) is Z+, therefore a|c QED