reserve p,q,r for FinSequence;
reserve u,v,x,y,y1,y2,z for object, A,D,X,Y for set;
reserve i,j,k,l,m,n for Nat;

theorem Th94:
  p is one-to-one & rng p = {x} implies len p = 1
proof
  assume that
A1: p is one-to-one and
A2: rng p = {x};
A3: now
    given y1,y2 such that
A4: y1 in dom p and
A5: y2 in dom p and
A6: y1 <> y2;
    p.y2 in rng p by A5,FUNCT_1:def 3;
    then
A7: p.y2 = x by A2,TARSKI:def 1;
    p.y1 in rng p by A4,FUNCT_1:def 3;
    then p.y1 = x by A2,TARSKI:def 1;
    hence contradiction by A1,A4,A5,A6,A7;
  end;
  set y = the Element of dom p;
A8: dom p <> {} by A2,RELAT_1:42;
A9: dom p = {y}
  proof
    thus dom p c= {y}
    proof
      let x be object;
      assume x in dom p;
      then x = y by A3;
      hence thesis by TARSKI:def 1;
    end;
    let x be object;
    y in dom p by A8;
    hence thesis by TARSKI:def 1;
  end;
  dom p = Seg(len p) by FINSEQ_1:def 3;
  hence thesis by A9,Th20;
end;
