let i, j, n, m be Element of NAT ; ( (abs (i - j)) + (abs (n - m)) = 1 iff ( ( abs (i - j) = 1 & n = m ) or ( abs (n - m) = 1 & i = j ) ) )
thus
( not (abs (i - j)) + (abs (n - m)) = 1 or ( abs (i - j) = 1 & n = m ) or ( abs (n - m) = 1 & i = j ) )
( ( ( abs (i - j) = 1 & n = m ) or ( abs (n - m) = 1 & i = j ) ) implies (abs (i - j)) + (abs (n - m)) = 1 )proof
assume A1:
(abs (i - j)) + (abs (n - m)) = 1
;
( ( abs (i - j) = 1 & n = m ) or ( abs (n - m) = 1 & i = j ) )
now per cases
( n = m or n <> m )
;
suppose A3:
n <> m
;
( ( abs (i - j) = 1 & n = m ) or ( abs (n - m) = 1 & i = j ) )now per cases
( n < m or n > m )
by A3, XXREAL_0:1;
suppose A4:
n < m
;
( ( abs (i - j) = 1 & n = m ) or ( abs (n - m) = 1 & i = j ) )then reconsider l =
m - n as
Element of
NAT by INT_1:18;
0 < l
by A4, XREAL_1:52;
then A5:
0 + 1
<= l
by NAT_1:13;
A6:
abs (- (m - n)) = abs l
by COMPLEX1:138;
then A7:
abs (n - m) = abs l
;
A8:
abs l = l
by ABSVALUE:def 1;
then A9:
l <= 1
by A1, A6, COMPLEX1:132, XREAL_1:33;
then
l = 1
by A5, XXREAL_0:1;
then
i - j = 0
by A1, A8, A7, ABSVALUE:7;
hence
( (
abs (i - j) = 1 &
n = m ) or (
abs (n - m) = 1 &
i = j ) )
by A6, A5, A8, A9, XXREAL_0:1;
verum end; end; end; hence
( (
abs (i - j) = 1 &
n = m ) or (
abs (n - m) = 1 &
i = j ) )
;
verum end; end; end;
hence
( (
abs (i - j) = 1 &
n = m ) or (
abs (n - m) = 1 &
i = j ) )
;
verum
end;
assume A14:
( ( abs (i - j) = 1 & n = m ) or ( abs (n - m) = 1 & i = j ) )
; (abs (i - j)) + (abs (n - m)) = 1
hence
(abs (i - j)) + (abs (n - m)) = 1
; verum