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
  for A,B,i st i in dom A & len A > 1 & Line(A,i) = width A |-> 0.K & i
in dom B & Line(B,i) = width B |-> 0.K holds Solutions_of(A,B) = Solutions_of(
  DelLine(A,i),DelLine(B,i))
proof
  let A,B,i such that
A1: i in dom A and
A2: len A > 1 and
A3: Line(A,i) = width A |-> 0.K and
A4: i in dom B and
A5: Line(B,i) = width B |-> 0.K;
  reconsider l1=len A-1 as Element of NAT by A2,NAT_1:20;
A6: l1 > 1-1 by A2,XREAL_1:9;
  thus Solutions_of(A,B) c= Solutions_of(DelLine(A,i),DelLine(B,i)) by A1,A2
,Th46;
  let x be object such that
A7: x in Solutions_of(DelLine(A,i),DelLine(B,i));
  set S=Seg len A;
A8: dom A=S by FINSEQ_1:def 3;
A9: now
    let j;
    assume j in (dom A)\(S\{i});
    then j in dom A /\{i} by A8,XBOOLE_1:48;
    then j in {i} by XBOOLE_0:def 4;
    hence Line(A,j) = width A |-> 0.K & Line(B,j) = width B |-> 0.K by A3,A5,
TARSKI:def 1;
  end;
  card S=l1+1 by FINSEQ_1:57;
  then
A10: card (S\{i})=l1 by A1,A8,STIRL2_1:55;
  (ex mA be Nat st len A = mA + 1 & len Del(A,i) = mA )& ex mB be Nat st
  len B = mB + 1 & len Del(B,i) = mB by A1,A4,FINSEQ_3:104;
  then
A11: len B=len A by A7,Th33;
  then dom A=dom B by A8,FINSEQ_1:def 3;
  then
  Solutions_of(A,B) = Solutions_of(Segm(A,S\{i},Seg width A),Segm(B,S\{i}
  ,Seg width B)) by A8,A10,A6,A9,Th45,CARD_1:27,XBOOLE_1:36
    .= Solutions_of(DelLine(A,i),Segm(B,S\{i},Seg width B)) by MATRIX13:51
    .= Solutions_of(DelLine(A,i),DelLine(B,i)) by A11,MATRIX13:51;
  hence thesis by A7;
end;
