let r, s be Real; ( |.(r - s).| = 1 iff ( ( r > s & r = s + 1 ) or ( r < s & s = r + 1 ) ) )
thus
( not |.(r - s).| = 1 or ( r > s & r = s + 1 ) or ( r < s & s = r + 1 ) )
( ( ( r > s & r = s + 1 ) or ( r < s & s = r + 1 ) ) implies |.(r - s).| = 1 )proof
assume A1:
|.(r - s).| = 1
;
( ( r > s & r = s + 1 ) or ( r < s & s = r + 1 ) )
now ( ( r > s & ( ( r > s & r = s + 1 ) or ( r < s & s = r + 1 ) ) ) or ( r = s & contradiction ) or ( r < s & ( ( r > s & r = s + 1 ) or ( r < s & s = r + 1 ) ) ) )end;
hence
( (
r > s &
r = s + 1 ) or (
r < s &
s = r + 1 ) )
;
verum
end;
assume A5:
( ( r > s & r = s + 1 ) or ( r < s & s = r + 1 ) )
; |.(r - s).| = 1