theorem Th36:
  [p1,p2] [= [q1,q2] iff p1 [= q1 & p2 [= q2
proof
  thus [p1,p2] [= [q1,q2] implies p1 [= q1 & p2 [= q2
  proof
    assume [p1,p2] "\/" [q1,q2] = [q1,q2];
    then [p1"\/"q1,p2"\/"q2] = [q1,q2] by Th21;
    hence p1"\/"q1 = q1 & p2"\/"q2 = q2 by XTUPLE_0:1;
  end;
  assume that
A1: p1"\/"q1 = q1 and
A2: p2"\/"q2 = q2;
  thus [p1,p2] "\/" [q1,q2] = [q1,q2] by A1,A2,Th21;
end;
