reserve x,y for set,
  i,j,k,l,m,n for Nat,
  K for Field,
  N for without_zero finite Subset of NAT,
  a,b for Element of K,
  A,B,B1,B2,X,X1,X2 for (Matrix of K),
  A9 for (Matrix of m,n,K),
  B9 for (Matrix of m,k,K);
reserve D for non empty set,
  bD for FinSequence of D,
  b,f,g for FinSequence of K,
  MD for Matrix of D;

theorem Th46:
  i in dom A & len A > 1 implies Solutions_of(A,B) c= Solutions_of
  (DelLine(A,i),DelLine(B,i))
proof
  assume that
A1: i in dom A and
A2: len A > 1;
  reconsider l1=len A-1 as Element of NAT by A2,NAT_1:20;
A3: l1 > 1-1 by A2,XREAL_1:9;
A4: Seg len A=dom A by FINSEQ_1:def 3;
  card Seg len A=l1+1 by FINSEQ_1:57;
  then card ((Seg len A)\{i})=l1 by A1,A4,STIRL2_1:55;
  then
A5: Solutions_of(A,B)c=Solutions_of(Segm(A,(Seg len A)\{i},Seg width A),
  Segm(B,(Seg len A)\{i},Seg width B)) by A4,A3,Th44,CARD_1:27,XBOOLE_1:36;
  let x be object such that
A6: x in Solutions_of(A,B);
  len A=len B by A6,Th33;
  then Segm(A,(Seg len A)\{i},Seg width A)=Del(A,i) & Segm(B,(Seg len A)\{i},
  Seg width B)=Del(B,i) by MATRIX13:51;
  hence thesis by A5,A6;
end;
