Are you talking about first order Peano Arithmetic? If so, this is usually a special case of one of the axioms.i want to construct a proof for 0 x 1 = 0 in first order system.

If your axioms have recursion on the second argument, as in the link above, then

0 x S(0) = 0 x 0 + 0 (by the second multiplication axiom)

= 0 + 0 (since 0 x 0 = 0 by the first multiplication axiom)

= 0 (by the first addition axiom)