# Reducing a boolean function with NAND |

The notation $$\ c|d \$$ is to be read 'c NAND d'

I have this function:

$$f(a,b,c,d)=(a+b)*(c|d)$$

$$=(a+b)*(\overline{c*d})$$ $$=(a+b)*(\bar{c}+\bar{d})$$ $$=(a\bar{c}+a\bar{d}+b\bar{c}+b\bar{d})$$ $$=(\bar{c}+\bar{d}+\bar{c}+\bar{d})$$ $$=(\bar{c}+\bar{d})=(\overline{cd})=(c|d)$$

There must be some mistakes I have done. My final result has to have a NAND between each two terms. Don't know how to get there.

• Create two truth tables, one from the original equation and one from your final result. If those truth tables have the same output then your two equations are equivalent. May 19, 2020 at 23:37
• My problem is the thing I have added. Each two terms in my end result has to have a NAND between them. May 19, 2020 at 23:40
• From your 3d equation : (a+b) * (not_c + not_d), in the next (4th) equation, i think you wrote * instead of + between the two terms May 20, 2020 at 0:34
• Ah yes! Now the states exercise does sense. Will edit my post. May 20, 2020 at 1:22
• should be correct now I guess?. Thanks! May 20, 2020 at 1:24