defpred S1[ Nat] means ex P, Q being finite without_zero Subset of NAT st
( [:P,Q:] c= Indices M & card P = card Q & card Q = $1 & Det (EqSegm M,P,Q) <> 0. K );
A1:
ex k being Nat st S1[k]
proof
consider E being
empty finite without_zero Subset of
NAT ;
reconsider E =
E as
finite without_zero Subset of
NAT ;
take C =
card E;
S1[C]
take P =
E;
ex Q being finite without_zero Subset of NAT st
( [:P,Q:] c= Indices M & card P = card Q & card Q = C & Det (EqSegm M,P,Q) <> 0. K )
take Q =
E;
( [:P,Q:] c= Indices M & card P = card Q & card Q = C & Det (EqSegm M,P,Q) <> 0. K )
A2:
E c= Seg (len M)
by XBOOLE_1:2;
A3:
E c= Seg (width M)
by XBOOLE_1:2;
Det (EqSegm M,E,E) = 1_ K
by CARD_1:47, MATRIXR2:41;
hence
(
[:P,Q:] c= Indices M &
card P = card Q &
card Q = C &
Det (EqSegm M,P,Q) <> 0. K )
by A2, A3, Th67;
verum
end;
A4:
for k being Nat st S1[k] holds
k <= len M
consider k being Nat such that
A9:
S1[k]
and
A10:
for n being Nat st S1[n] holds
n <= k
from NAT_1:sch 6(A4, A1);
take
k
; ( k is Element of NAT & ex P, Q being finite without_zero Subset of NAT st
( [:P,Q:] c= Indices M & card P = card Q & card P = k & Det (EqSegm M,P,Q) <> 0. K ) & ( for P1, Q1 being finite without_zero Subset of NAT st [:P1,Q1:] c= Indices M & card P1 = card Q1 & Det (EqSegm M,P1,Q1) <> 0. K holds
card P1 <= k ) )
thus
( k is Element of NAT & ex P, Q being finite without_zero Subset of NAT st
( [:P,Q:] c= Indices M & card P = card Q & card P = k & Det (EqSegm M,P,Q) <> 0. K ) & ( for P1, Q1 being finite without_zero Subset of NAT st [:P1,Q1:] c= Indices M & card P1 = card Q1 & Det (EqSegm M,P1,Q1) <> 0. K holds
card P1 <= k ) )
by A9, A10; verum