
theorem Th4:
  for M being non void non empty SubsetFamilyStr holds M is
with_exchange_property iff for A,B being finite Subset of M st A is independent
& B is independent & card B = (card A) + 1 ex e being Element of M st e in B \
  A & A \/ {e} is independent
proof
  let M be non void non empty SubsetFamilyStr;

  thus M is with_exchange_property implies for A,B being finite Subset of M st
A is independent & B is independent & card B = (card A) + 1 ex e being Element
  of M st e in B \ A & A \/ {e} is independent
  proof
    assume
A1: for A,B being finite Subset of M st A in the_family_of M & B in
the_family_of M & card B = (card A) + 1 ex e being Element of M st e in B \ A &
    A \/ {e} in the_family_of M;
    let A,B be finite Subset of M;
    assume that
A2: A in the_family_of M and
A3: B in the_family_of M and
A4: card B = (card A) + 1;
    consider e being Element of M such that
A5: e in B \ A and
A6: A \/ {e} in the_family_of M by A1,A2,A3,A4;
    take e;
    thus e in B \ A & A \/ {e} in the_family_of M by A5,A6;
  end;
  assume
A7: for A,B being finite Subset of M st A is independent & B is
  independent & card B = (card A) + 1 ex e being Element of M st e in B \ A & A
  \/ {e} is independent;
  let A,B be finite Subset of M;
  assume that
A8: A in the_family_of M and
A9: B in the_family_of M;
A10: B is independent by A9;
  assume
A11: card B = (card A) + 1;
  A is independent by A8;
  then consider e being Element of M such that
A12: e in B \ A and
A13: A \/ {e} is independent by A7,A10,A11;
  take e;
  thus thesis by A12,A13;
end;
