let K be Field; for M1, M2 being Matrix of K st len M1 = len M2 & width M1 = width M2 & M1 - M2 = 0. (K,(len M1),(width M1)) holds
M1 = M2
let M1, M2 be Matrix of K; ( len M1 = len M2 & width M1 = width M2 & M1 - M2 = 0. (K,(len M1),(width M1)) implies M1 = M2 )
assume that
A1:
len M1 = len M2
and
A2:
width M1 = width M2
and
A3:
M1 - M2 = 0. (K,(len M1),(width M1))
; M1 = M2
per cases
( len M1 > 0 or len M1 = 0 )
by NAT_1:3;
suppose A4:
len M1 > 0
;
M1 = M2then A5:
M2 is
Matrix of
len M1,
width M1,
K
by A1, A2, MATRIX_0:20;
A6:
(
len (- M2) = len M2 &
width (- M2) = width M2 )
by MATRIX_3:def 2;
A7:
len (0. (K,(len M1),(width M1))) = len M1
by MATRIX_0:def 2;
then
width (0. (K,(len M1),(width M1))) = width M1
by A4, MATRIX_0:20;
then (M1 + (- M2)) + M2 =
M2 + (0. (K,(len M1),(width M1)))
by A1, A2, A3, A7, MATRIX_3:2
.=
M2
by A5, MATRIX_3:4
;
then
M1 + ((- M2) + M2) = M2
by A1, A2, A6, MATRIX_3:3;
then
M1 + (M2 + (- M2)) = M2
by A6, MATRIX_3:2;
then A8:
M1 + (0. (K,(len M1),(width M1))) = M2
by A5, MATRIX_3:5;
M1 is
Matrix of
len M1,
width M1,
K
by A4, MATRIX_0:20;
hence
M1 = M2
by A8, MATRIX_3:4;
verum end; end;