# simplify the boolean expression $abc|ab\sim c|a\sim bc|\sim abc$

I worked this through to a&c but this has to be wrong. I'm clearly going wrong somewhere. Could someone point out the wrong step in my method?

$$(a\land b\land c)\lor (a\land b\land \lnot c)\lor (a\land \lnot b\land c)\lor (\lnot a\land b\land c)$$ if $d=b\land c$ and $e = b \land \lnot c$ then by substitution :

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

then by distribution:

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

then by substituting $d$ and $e$ back : $$a\land ((b\land c)\lor(b\land \lnot c))\lor (a\land \lnot b\land c)\lor (\lnot a\land b\land c)$$

then by distribution for b: $$a\land (b\land (c\lor \lnot c))\lor (a\land \lnot b\land c)\lor (\lnot a\land b\land c)$$

inversion of c: $$a\land (b\land 1)\lor (a\land \lnot b\land c)\lor (\lnot a\land b\land c)$$ identity for b: $$a\land b\lor (a\land \lnot b\land c)\lor (\lnot a\land b\land c)$$

= a & b | a&~b&c | ~a&b&c

f = a&~b&c

= f | a & b | ~a&b&c

= f&a | f&b | ~a&b&c

= a&~b&c&a | a&~b&c&b | ~a&b&c

= a&~b&c | ~a&b&c

g = ~b&c , h =b&c

= a&g | ~a&h

= a&(a|~a)&g | ~a&h

= a&(a&g|~a&g) | ~a&h

= a&g|~a&(a&g) | ~a&h

= a&g|~a&((a&g)|h)

= g & a | (~a) & ((a&g)|h)

= g & (a & ~a) | a & ((a&g)|h)

= g & (a & ~a) | a & ((a&g)|h)

= g & 0 | a&(a&g) | a&h

= a&g | a&h

= a&~b&c | a&b&c

= (a&c) & (~b|b)

= a&c&1

= a&c

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You are at least wrong at:

f = a&~b&c
f | a & b | ~a&b&c = f&a | f&b | ~a&b&c


or should I write:

$$f = a\land \neg b \land c$$ $$f \lor (a \land b) \lor (\neg a \land b \land c) = (f \land a) \lor (f \land b) \lor (\neg a \land b \land c)$$

$f \land b$ expands to $a \land \neg b \land c \land b \neq a \land b$. To simplify your expression observe that it is symmetric in terms of $a, b, c$ and it is true if and only if no less than 2 of them are true, so the simplification could be: $(a\land b) \lor (b \land c) \lor (c \land a)$ or $(a \land (b \lor c)) \lor (b \land c)$ or similar things.

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To simplify this expression you can use Karnaugh map :

Hence :

$f = (a \land b) \lor ( a \land c) \lor (b \land c)$

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