theorem Th6:
  <:f,g:>|A = <:f,g|A:>
proof
  thus <:f,g:>|A = <:g,f:>~|A by Th2
    .= (<:g,f:>|A)~ by Th3
    .= <:g|A,f:>~ by Th5
    .= <:f,g|A:> by Th2;
end;
