let K be Field; :: thesis: for A, B being Matrix of K
for i being Nat st i in dom A & len A > 1 & Line (A,i) = () |-> (0. K) & i in dom B & Line (B,i) = () |-> (0. K) holds
Solutions_of (A,B) = Solutions_of ((DelLine (A,i)),(DelLine (B,i)))

let A, B be Matrix of K; :: thesis: for i being Nat st i in dom A & len A > 1 & Line (A,i) = () |-> (0. K) & i in dom B & Line (B,i) = () |-> (0. K) holds
Solutions_of (A,B) = Solutions_of ((DelLine (A,i)),(DelLine (B,i)))

let i be Nat; :: thesis: ( i in dom A & len A > 1 & Line (A,i) = () |-> (0. K) & i in dom B & Line (B,i) = () |-> (0. K) implies Solutions_of (A,B) = Solutions_of ((DelLine (A,i)),(DelLine (B,i))) )
assume that
A1: i in dom A and
A2: len A > 1 and
A3: Line (A,i) = () |-> (0. K) and
A4: i in dom B and
A5: Line (B,i) = () |-> (0. K) ; :: thesis: Solutions_of (A,B) = Solutions_of ((DelLine (A,i)),(DelLine (B,i)))
reconsider l1 = (len A) - 1 as Element of NAT by ;
A6: l1 > 1 - 1 by ;
thus Solutions_of (A,B) c= Solutions_of ((DelLine (A,i)),(DelLine (B,i))) by A1, A2, Th46; :: according to XBOOLE_0:def 10 :: thesis: Solutions_of ((DelLine (A,i)),(DelLine (B,i))) c= Solutions_of (A,B)
let x be object ; :: according to TARSKI:def 3 :: thesis: ( not x in Solutions_of ((DelLine (A,i)),(DelLine (B,i))) or x in Solutions_of (A,B) )
assume A7: x in Solutions_of ((DelLine (A,i)),(DelLine (B,i))) ; :: thesis: x in Solutions_of (A,B)
set S = Seg (len A);
A8: dom A = Seg (len A) by FINSEQ_1:def 3;
A9: now :: thesis: for j being Nat st j in (dom A) \ ((Seg (len A)) \ {i}) holds
( Line (A,j) = () |-> (0. K) & Line (B,j) = () |-> (0. K) )
let j be Nat; :: thesis: ( j in (dom A) \ ((Seg (len A)) \ {i}) implies ( Line (A,j) = () |-> (0. K) & Line (B,j) = () |-> (0. K) ) )
assume j in (dom A) \ ((Seg (len A)) \ {i}) ; :: thesis: ( Line (A,j) = () |-> (0. K) & Line (B,j) = () |-> (0. K) )
then j in (dom A) /\ {i} by ;
then j in {i} by XBOOLE_0:def 4;
hence ( Line (A,j) = () |-> (0. K) & Line (B,j) = () |-> (0. K) ) by ; :: thesis: verum
end;
card (Seg (len A)) = l1 + 1 by FINSEQ_1:57;
then A10: card ((Seg (len A)) \ {i}) = l1 by ;
( ex mA being Nat st
( len A = mA + 1 & len (Del (A,i)) = mA ) & ex mB being Nat st
( len B = mB + 1 & len (Del (B,i)) = mB ) ) by ;
then A11: len B = len A by ;
then dom A = dom B by ;
then Solutions_of (A,B) = Solutions_of ((Segm (A,((Seg (len A)) \ {i}),(Seg ()))),(Segm (B,((Seg (len A)) \ {i}),(Seg ())))) by
.= Solutions_of ((DelLine (A,i)),(Segm (B,((Seg (len A)) \ {i}),(Seg ())))) by MATRIX13:51
.= Solutions_of ((DelLine (A,i)),(DelLine (B,i))) by ;
hence x in Solutions_of (A,B) by A7; :: thesis: verum