theorem Th35:
  A * B * C = A * (B + C)
proof
  thus A * B * C c= A * (B + C)
  proof
    let x be object;
    assume x in A * B * C;
    then consider a,c such that
A1: x = a * c and
A2: a in A * B and
A3: c in C;
    consider g,h such that
A4: a = g * h and
A5: g in A and
A6: h in B by A2;
    x = g * (h + c) & h + c in B + C by A1,A3,A4,A6,Th24;
    hence thesis by A5;
  end;
  let x be object;
  assume x in A * (B + C);
  then consider a,b such that
A7: x = a * b & a in A and
A8: b in B + C;
  consider g,h such that
A9: b = g + h & g in B and
A10: h in C by A8;
  a * g in A * B & x = a * g * h by A7,A9,Th24;
  hence thesis by A10;
end;
