let K be Field; :: thesis: for M1, M2 being Matrix of K st len M1 = len M2 & width M1 = width M2 & M2 - M1 = M2 holds
M1 = 0. (K,(len M1),(width M1))

let M1, M2 be Matrix of K; :: thesis: ( len M1 = len M2 & width M1 = width M2 & M2 - M1 = M2 implies M1 = 0. (K,(len M1),(width M1)) )
assume that
A1: ( len M1 = len M2 & width M1 = width M2 ) and
A2: M2 - M1 = M2 ; :: thesis: M1 = 0. (K,(len M1),(width M1))
per cases ( len M1 > 0 or len M1 = 0 ) by NAT_1:3;
suppose A3: len M1 > 0 ; :: thesis: M1 = 0. (K,(len M1),(width M1))
A4: ( len (- M1) = len M1 & width (- M1) = width M1 ) by MATRIX_3:def 2;
A5: M2 is Matrix of len M1, width M1,K by ;
then (M2 + (- M1)) + (- M2) = 0. (K,(len M1),(width M1)) by ;
then ((- M1) + M2) + (- M2) = 0. (K,(len M1),(width M1)) by ;
then (- M1) + (M2 + (- M2)) = 0. (K,(len M1),(width M1)) by ;
then A6: (- M1) + (0. (K,(len M1),(width M1))) = 0. (K,(len M1),(width M1)) by ;
- M1 is Matrix of len M1, width M1,K by ;
then - M1 = 0. (K,(len M1),(width M1)) by ;
then M1 = - (0. (K,(len M1),(width M1))) by Th1;
hence M1 = 0. (K,(len M1),(width M1)) by Th9; :: thesis: verum
end;
suppose A7: len M1 = 0 ; :: thesis: M1 = 0. (K,(len M1),(width M1))
then len (0. (K,(len M1),(width M1))) = 0 ;
hence M1 = 0. (K,(len M1),(width M1)) by ; :: thesis: verum
end;
end;