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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.

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    \$\begingroup\$ 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. \$\endgroup\$ May 19, 2020 at 23:37
  • \$\begingroup\$ My problem is the thing I have added. Each two terms in my end result has to have a NAND between them. \$\endgroup\$
    – Rapiz
    May 19, 2020 at 23:40
  • \$\begingroup\$ 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 \$\endgroup\$
    – Atomique
    May 20, 2020 at 0:34
  • \$\begingroup\$ Ah yes! Now the states exercise does sense. Will edit my post. \$\endgroup\$
    – Rapiz
    May 20, 2020 at 1:22
  • \$\begingroup\$ should be correct now I guess?. Thanks! \$\endgroup\$
    – Rapiz
    May 20, 2020 at 1:24

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