Symmetric chain decomposition of cartesian product 
Let $A,B$ be finite posets. Define '$\leq$' for the cartesian product
  $A\times B$ as $(a,b) \leq (a',b') \Leftrightarrow a \leq a' \wedge  b
  \leq b' \forall a \in A, b \in B $. 
Show that there is a symmetric chain decomposition (partition into
  symmetric chains) for $A\times B$ if there is any for $A$ and $B$.

I played around with some examples, but didn't find any beginning for a proof. Could you please help me to go on?

Do this in a few steps:
(1) Take the assumed symmetric chain decompositions of A and B, say A_1, A_2, . . . , A_s for
A and B_1, B_2, . . . , B_t for B and create the s x t chain products A_i x B_j ;
(2) Argue that each of the chain products is a "symmetric" and "saturated" subposet of A x B;
(3) Prove that a product of two chains has a symmetric chain decomposition;
(4) Argue that the union of these SCD's provides an SCD of A x B.
