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