reserve k,n for Nat,
  x,y,z,y1,y2 for object,X,Y for set,
  f,g for Function;
reserve p,q,r,s,t for XFinSequence;
reserve D for set;

theorem Th35:
  p = <%x,y%> iff len p = 2 & p.0=x & p.1=y
proof
  thus p = <%x,y%> implies len p = 2 & p.0=x & p.1=y
  proof
    assume
A1: p=<%x,y%>;
    hence len p = len <%x%> + len <%y%> by Def3
      .= 1 + len <%y%> by Th30
      .= 1 + 1 by Th30
      .=2;
    0 in {0} by TARSKI:def 1;
    then
A3: 0 in dom <%y%>;
    0 in dom <%x%> by TARSKI:def 1;
    hence p.0 = <%x%>.0 by A1,Def3
      .= x;
    thus p.1 = (<%x%>^<%y%>).(len <%x%> + 0) by A1,Th30
      .= <%y%>.0 by A3,Def3
      .= y;
  end;
  assume that
A4: len p = 2 and
A5: p.0=x and
A6: p.1=y;
A7: for k st k in dom <%y%> holds p.((len <%x%>)+k)=<%y%>.k
  proof
    let k;
    assume a8: k in dom <%y%>;
    thus p.((len <%x%>) + k) = p.(1+k) by Th30
      .=p.(1+0) by a8,TARSKI:def 1
      .=<%y%>.0 by A6
      .= <%y%>.k by a8,TARSKI:def 1;
  end;
A9: for k st k in dom <%x%> holds p.k=<%x%>.k
  proof
    let k;
    assume k in dom <%x%>;
    then k=0 by TARSKI:def 1;
    hence thesis by A5;
  end;
  dom p = (1+1) by A4
    .= (len <%x%> + 1) by Th30
    .= (len <%x%> + len <%y%>) by Th30;
  hence thesis by A9,A7,Def3;
end;
