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$\begingroup$

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

2 Answers 2