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
set E = the
empty finite without_zero Subset of
NAT;
reconsider E = the
empty finite without_zero Subset of
NAT as
finite without_zero Subset of
NAT ;
take
card E
;
S1[ card E]
take
E
;
ex Q being finite without_zero Subset of NAT st
( [:E,Q:] c= Indices M & card E = card Q & card Q = card E & Det (EqSegm (M,E,Q)) <> 0. K )
take
E
;
( [:E,E:] c= Indices M & card E = card E & card E = card E & Det (EqSegm (M,E,E)) <> 0. K )
A2:
E c= Seg (len M)
;
A3:
E c= Seg (width M)
;
Det (EqSegm (M,E,E)) = 1_ K
by MATRIXR2:41;
hence
(
[:E,E:] c= Indices M &
card E = card E &
card E = card E &
Det (EqSegm (M,E,E)) <> 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