let K be Field; :: thesis: for M being Matrix of K
for P, P1, Q1 being finite without_zero Subset of NAT st card P1 = card Q1 & P c= P1 & Det (EqSegm (M,P1,Q1)) <> 0. K holds
ex Q being finite without_zero Subset of NAT st
( Q c= Q1 & card P = card Q & Det (EqSegm (M,P,Q)) <> 0. K )

let M be Matrix of K; :: thesis: for P, P1, Q1 being finite without_zero Subset of NAT st card P1 = card Q1 & P c= P1 & Det (EqSegm (M,P1,Q1)) <> 0. K holds
ex Q being finite without_zero Subset of NAT st
( Q c= Q1 & card P = card Q & Det (EqSegm (M,P,Q)) <> 0. K )

let P, P1, Q1 be finite without_zero Subset of NAT; :: thesis: ( card P1 = card Q1 & P c= P1 & Det (EqSegm (M,P1,Q1)) <> 0. K implies ex Q being finite without_zero Subset of NAT st
( Q c= Q1 & card P = card Q & Det (EqSegm (M,P,Q)) <> 0. K ) )

assume that
A1: card P1 = card Q1 and
A2: P c= P1 and
A3: Det (EqSegm (M,P1,Q1)) <> 0. K ; :: thesis: ex Q being finite without_zero Subset of NAT st
( Q c= Q1 & card P = card Q & Det (EqSegm (M,P,Q)) <> 0. K )

defpred S1[ Nat] means for P being finite without_zero Subset of NAT st P c= P1 & card P = $1 holds
ex Q being finite without_zero Subset of NAT st
( Q c= Q1 & card P = card Q & Det (EqSegm (M,P,Q)) <> 0. K );
A4: for k being Element of NAT st k < card P1 & S1[k + 1] holds
S1[k]
proof
let k be Element of NAT ; :: thesis: ( k < card P1 & S1[k + 1] implies S1[k] )
assume that
A5: k < card P1 and
A6: S1[k + 1] ; :: thesis: S1[k]
let P be finite without_zero Subset of NAT; :: thesis: ( P c= P1 & card P = k implies ex Q being finite without_zero Subset of NAT st
( Q c= Q1 & card P = card Q & Det (EqSegm (M,P,Q)) <> 0. K ) )

assume that
A7: P c= P1 and
A8: card P = k ; :: thesis: ex Q being finite without_zero Subset of NAT st
( Q c= Q1 & card P = card Q & Det (EqSegm (M,P,Q)) <> 0. K )

P c< P1 by A5, A7, A8, XBOOLE_0:def 8;
then P1 \ P <> {} by XBOOLE_1:105;
then consider x being object such that
A9: x in P1 \ P by XBOOLE_0:def 1;
reconsider x = x as non zero Element of NAT by A9;
reconsider Px = P \/ {x} as finite without_zero Subset of NAT ;
A10: not x in P by A9, XBOOLE_0:def 5;
then A11: card Px = k + 1 by A8, CARD_2:41;
x in P1 by A9, XBOOLE_0:def 5;
then {x} c= P1 by ZFMISC_1:31;
then Px c= P1 by A7, XBOOLE_1:8;
then consider Q2 being finite without_zero Subset of NAT such that
A12: Q2 c= Q1 and
A13: card Px = card Q2 and
A14: Det (EqSegm (M,Px,Q2)) <> 0. K by A6, A8, A10, CARD_2:41;
set E = EqSegm (M,Px,Q2);
A15: Px \ {x} = P by A10, ZFMISC_1:117;
x in {x} by TARSKI:def 1;
then A16: x in Px by XBOOLE_0:def 3;
A18: dom (Sgm Px) = Seg (card Px) by FINSEQ_3:40;
rng (Sgm Px) = Px by FINSEQ_1:def 14;
then consider i being object such that
A19: i in Seg (card Px) and
A20: (Sgm Px) . i = x by A18, A16, FUNCT_1:def 3;
A21: (k + 1) -' 1 = (k + 1) - 1 by XREAL_0:def 2;
reconsider i = i as Element of NAT by A19;
consider j being Nat such that
A22: j in Seg (card Px) and
A23: Det (Delete ((EqSegm (M,Px,Q2)),i,j)) <> 0. K by A14, A19, Lm3;
take Q = Q2 \ {((Sgm Q2) . j)}; :: thesis: ( Q c= Q1 & card P = card Q & Det (EqSegm (M,P,Q)) <> 0. K )
Q c= Q2 by XBOOLE_1:36;
hence ( Q c= Q1 & card P = card Q & Det (EqSegm (M,P,Q)) <> 0. K ) by A8, A11, A12, A13, A19, A20, A22, A23, A15, A21, Th64; :: thesis: verum
end;
A24: S1[ card P1]
proof
let P be finite without_zero Subset of NAT; :: thesis: ( P c= P1 & card P = card P1 implies ex Q being finite without_zero Subset of NAT st
( Q c= Q1 & card P = card Q & Det (EqSegm (M,P,Q)) <> 0. K ) )

assume that
A25: P c= P1 and
A26: card P = card P1 ; :: thesis: ex Q being finite without_zero Subset of NAT st
( Q c= Q1 & card P = card Q & Det (EqSegm (M,P,Q)) <> 0. K )

P = P1 by A25, A26, CARD_2:102;
hence ex Q being finite without_zero Subset of NAT st
( Q c= Q1 & card P = card Q & Det (EqSegm (M,P,Q)) <> 0. K ) by A1, A3; :: thesis: verum
end;
for k being Element of NAT st k <= card P1 holds
S1[k] from PRE_POLY:sch 1(A24, A4);
hence ex Q being finite without_zero Subset of NAT st
( Q c= Q1 & card P = card Q & Det (EqSegm (M,P,Q)) <> 0. K ) by A2, NAT_1:43; :: thesis: verum