theorem Th75:
  [:rng nt1,rng nt2:] c= Indices M & Det Segm(M,nt1,nt2)<>0.K
implies ex P1,P2 st P1 = rng nt1 & P2 = rng nt2 & card P1 = card P2 & card P1 =
  n & Det EqSegm(M,P1,P2) <>0.K
proof
  assume that
A1: [:rng nt1,rng nt2:] c= Indices M and
A2: Det Segm(M,nt1,nt2)<>0.K;
  n=0 iff n=0;
  then consider P1,P2 such that
A3: P1 = rng nt1 and
A4: P2 = rng nt2 by A1,Lm5;
  nt2 is one-to-one by A2,Th31;
  then
A5: card P2 =len nt2 by A4,FINSEQ_4:62;
  nt1 is one-to-one by A2,Th27;
  then
A6: card P1 =len nt1 by A3,FINSEQ_4:62;
  then reconsider SP1=Sgm P1,SP2=Sgm P2 as Element of n-tuples_on NAT by A5,
CARD_1:def 7;
A7: rng SP2=P2 by FINSEQ_1:def 14;
  rng SP1=P1 by FINSEQ_1:def 14;
  then
A8: Det Segm(M,nt1,nt2) = Det Segm(M,SP1,SP2) or -Det Segm(M,nt1,nt2) = Det
  Segm(M,SP1,SP2) by A3,A4,A7,Th36;
A9: len nt1=n by CARD_1:def 7;
A10: len nt2=n by CARD_1:def 7;
  Segm(M,Sgm P1,Sgm P2) = Segm(M,P1,P2)
    .= EqSegm(M,P1,P2) by A6,A5,A9,A10,Def3;
  hence thesis by A2,A3,A4,A6,A5,A9,A10,A8,VECTSP_1:28;
end;
