theorem Th62:
  p is one-to-one iff card rng p = len p
proof
  thus p is one-to-one implies card rng p = len p
  proof
    assume p is one-to-one;
    then dom p,p .: (dom p) are_equipotent by CARD_1:33;
    then dom p,rng p are_equipotent by RELAT_1:113;
    then Seg len p, rng p are_equipotent by FINSEQ_1:def 3;
    then card Seg len p = card rng p by CARD_1:5;
    hence thesis by FINSEQ_1:57;
  end;
  reconsider f = p as Function of dom p, rng p by FUNCT_2:1;
  reconsider B = dom p as finite set;
  assume card rng p = len p;
  then card rng p = card Seg len p by FINSEQ_1:57;
  then
A1: card rng p = card B by FINSEQ_1:def 3;
  f is onto by FUNCT_2:def 3;
  hence thesis by Lm1,A1;
end;
