reserve x,y for object,X,Y for set,
  D for non empty set,
  i,j,k,l,m,n,m9,n9 for Nat,
  i0,j0,n0,m0 for non zero Nat,
  K for Field,
  a,b for Element of K,
  p for FinSequence of K,
  M for Matrix of n,K;
reserve A for (Matrix of D),
  A9 for Matrix of n9,m9,D,
  M9 for Matrix of n9, m9,K,
  nt,nt1,nt2 for Element of n-tuples_on NAT,
  mt,mt1 for Element of m -tuples_on NAT,
  M for Matrix of K;
reserve P,P1,P2,Q,Q1,Q2 for without_zero finite Subset of NAT;

theorem Th80:
  P c= P1 & Q c= Q1 implies the_rank_of Segm(M,P,Q) <= the_rank_of
  Segm(M,P1,Q1)
proof
  assume that
A1: P c= P1 and
A2: Q c= Q1;
  set S1=Segm(M,P1,Q1);
  set S=Segm(M,P,Q);
  consider P2,Q2 such that
A3: [:P2,Q2:] c= Indices S and
A4: card P2 = card Q2 and
A5: card P2 = the_rank_of S and
A6: Det EqSegm(S,P2,Q2)<>0.K by Def4;
  P2={} iff Q2={} by A4;
  then consider P3,Q3 be without_zero finite Subset of NAT such that
A7: P3 c= P and
A8: Q3 c= Q and
  P3 = Sgm P.:P2 and
  Q3=Sgm Q.:Q2 and
A9: card P3=card P2 and
A10: card Q3=card Q2 and
A11: Segm(S,P2,Q2) = Segm(M,P3,Q3) by A3,Th57;
  reconsider P4=Sgm P1"P3, Q4=Sgm Q1"Q3 as without_zero finite Subset of NAT
  by Th53;
A12: card Q4=card P2 by A2,A4,A8,A10,Lm2,XBOOLE_1:1;
A13: card P4=card P2 by A1,A7,A9,Lm2,XBOOLE_1:1;
A14: rng Sgm Q4=Q4 by FINSEQ_1:def 14;
A15: Q3 c= Q1 by A2,A8;
A16: P3 c= P1 by A1,A7;
  rng Sgm P4=P4 by FINSEQ_1:def 14;
  then
A17: [:P4,Q4:] c= Indices S1 by A16,A15,A14,Th56;
  Segm(S1,P4,Q4) = Segm(M,P3,Q3) by A16,A15,Th56;
  then EqSegm(S,P2,Q2) = Segm(S1,P4,Q4) by A4,A11,Def3
    .= EqSegm(S1,P4,Q4) by A4,A10,A15,A13,Def3,Lm2;
  hence thesis by A5,A6,A13,A12,A17,Def4;
end;
