Transistor count in NAND or NOR implementation of boolean algebra?
Your logic looks like a sum of products (ANDs feeding into an OR). In that case, you can replace both the ANDs and the OR with NAND gates. Also, you can replace (~a ~c) with ~(a + c), which saves a gate. Putting it together, that gives 1
NOR, 1 NOT, and 2 NAND, for a total of 14 transistors.