How to simplify boolean algebra (~a~b)+(abc)+(a~c)

Question

I've been trying for ages and keep getting (~a~b)+(ab)+(a~c), but all simplification websites keep saying (~a~b)+(ab)+(~b~c).

My reasoning:

• (~a~b)+(abc)+(a~c)
• (~a~b)+(a(bc+~c))
• (~a~b)+(a((b+~c)(c+~c)))
• (~a~b)+(a((b+~c)(1)))
• (~a~b)+(a(b+~c))
• (~a~b)+(ab)+(a~c)

Any help would be greatly appreciated, thanks.

~ = NOT

ab = a AND b

a+b = a OR b

Answer 1

Both answers are correct, as they are equivalent:

$$a'b'+ab+ac'$$

$$\overset{Adjacency}{=}$$

$$(a'b'c+a'b'c')+(abc+abc')+(abc'+ab'c')$$

$$\overset{Association}{=}$$

$$a'b'c+a'b'c'+abc+abc'+abc'+ab'c'$$

$$\overset{Idempotence}{=}$$

$$a'b'c+a'b'c'+abc+abc'+ab'c'$$

$$\overset{Idempotence}{=}$$

$$a'b'c+a'b'c'+a'b'c'+abc+abc'+ab'c'$$

$$\overset{Commutation}{=}$$

$$a'b'c+a'b'c'+abc+abc'+ab'c'+a'b'c'$$

$$\overset{Association}{=}$$

$$(a'b'c+a'b'c')+(abc+abc')+(ab'c'+a'b'c')$$

$$\overset{Adjacency}{=}$$

$$a'b'+ab+b'c'$$