reserve i,j,m,n,k for Nat,
  x,y for set,
  K for Field,
  a for Element of K;

theorem Th6:
  i in Seg n implies Delete(1.(K,n),i,i)=1.(K,n-'1)
proof
  assume
A1: i in Seg n;
  then n <> 0;
  then n >= 1 by NAT_1:14;
  then n-'1=n-1 by XREAL_1:233;
  then card Seg n=n-'1+1 by FINSEQ_1:57;
  then
A2: card (Seg n\{i})=n-'1 by A1,STIRL2_1:55;
  thus Delete(1.(K,n),i,i) = Deleting(1.(K,n),i,i) by A1,LAPLACE:def 1
    .= Segm(1.(K,n),Seg n\{i},Seg n\{i}) by MATRIX13:58
    .= 1.(K,n-'1) by A2,Th2,XBOOLE_1:36;
end;
