let K be Ring; :: thesis: 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; :: thesis: ( 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)) ; :: thesis: M1 = M2

per cases ( len M1 > 0 or len M1 = 0 ) by NAT_1:3;

suppose A4: len M1 > 0 ; :: thesis: M1 = M2

then 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; :: thesis: verum

end;

suppose len M1 = 0 ; :: thesis: M1 = M2

hence M1 = M2 by A1, CARD_2:64; :: thesis: verum

end;

end;