Simplify this Logic Function?

Question

Have a Hardware Lab to do, and I need to reduce the following function before I actually hook it up to the Logic Trainer.

(not ac) + (abc) + (a not c)

Or:

$\lnot (a \land c) \lor (a \land b \land c) \lor (a \land \lnot c)$

Thanks!

Answer 1

I think this is a reduction: $\neg(a\wedge c)$ equals to $(a\wedge\neg c)\vee(\neg a\wedge c)\vee(\neg a\wedge\neg c)$. But $(a\wedge\neg c)$ already exists at your formula, so you can just ignore it and have $\neg(a\wedge c)\vee(a\wedge b\wedge c)$

Answer 2

¬(a∧c) = ¬(a∧c)

so ¬(a∧c)∨(a∧b∧c)∨(a∧¬c) = ¬(a∧c)∨(a∧b∧c)

for a∧c = x

-x ∨ (x∧b)

---

which means

if b = 1 answer is -x

if b = 0 answer is 0

= b∨-x

in other words

-x ∨ (x∧b) = (-x∨x) ∧ (-x∨b)

---

= 0 ∧ (-x∨b)

= -x∨b

= ¬(a∧c)∨b

Answer 3

$\neg(a \wedge c)\lor(a \wedge b \wedge c)$ is not equivalent to the formula in question. The assignment $v$ with $v(a)=v(c)=1, v(b)=0~$ falsifies $\neg(a \wedge c)\lor(a \wedge b \wedge c)$, but verifies the original formula.