reserve i,j,n for Nat,
  K for Field,
  a for Element of K,
  M,M1,M2,M3,M4 for Matrix of n,K;
reserve A for Matrix of K;

theorem
  for K being Ring, n being Nat, M1,M2,M3,M4 being Matrix of n,K st
  M3 is_reverse_of M1 & M4 is_reverse_of M2 holds M3*M4 is_reverse_of M2*M1
proof
  let K be Ring,n be Nat,M1,M2,M3,M4 be Matrix of n,K;
A1: width M1=n by MATRIX_0:24;
A2: width M2=n & len M1=n by MATRIX_0:24;
A3: len M3=n by MATRIX_0:24;
A4: width M3=n & len M4=n by MATRIX_0:24;
  assume that
A5: M3 is_reverse_of M1 and
A6: M4 is_reverse_of M2;
  width (M2*M1)=n by MATRIX_0:24;
  then
A7: (M2*M1)*(M3*M4)=((M2*M1)*M3)*M4 by A3,A4,MATRIX_3:33
    .=(M2*(M1*M3))*M4 by A1,A2,A3,MATRIX_3:33
    .=(M2*(1.(K,n)))*M4 by A5
    .=M2*M4 by MATRIX_3:19
    .=1.(K,n) by A6;
A8: width M4=n by MATRIX_0:24;
A9: len M2=n by MATRIX_0:24;
  width (M3*M4)=n by MATRIX_0:24;
  then (M3*M4)*(M2*M1)=((M3*M4)*M2)*M1 by A2,A9,MATRIX_3:33
    .=(M3*(M4*M2))*M1 by A9,A8,A4,MATRIX_3:33
    .=(M3*(1.(K,n)))*M1 by A6
    .=M3*M1 by MATRIX_3:19
    .=1.(K,n) by A5;
  hence thesis by A7;
end;
