Answer to Question #217671 in Discrete Mathematics for Kunal

Question

Produce a truth table for given Boolean expression (A+B'+C)(A+B+C)(A'+B+C')

Answer 1

$(A+B'+C)(A+B+C)(A'+B+C')$

Laws Used:

$A\cdot A =A\\A+AB=A\ \ [\text{Absorption Law}]$

$A\cdot A'=0\\A+0=A\\A+A=A\\A+1=1$

Now ,

$\implies (A+B'+C)(A+B+C)(A'+B+C')\\\implies (A+AB+AC+AB'+B'C+AC+BC+C)(A'+B+C')\\\implies (AA'+AB+AC'+AA'B+ABB+ABC'+AA'C+ABC+ACC'+AA'B'+ABB'+AB'C'+A'B'C+BB'C+B'CC'+AA'C+ABC+ACC'+A'BC+BBC+BCC'+A'C+BC+CC')$

$\implies (0+AB+AC'+0+AB+ABC'+0+ABC+0+0+0+AB'C'+A'B'C+0+0+0+ABC+0+A'BC+BC+0+A'C+BC+0)\\\implies (AB+BC+AC+A'C+ABC+ABC'+AB'C'+A'B'C+A'BC)\\\implies AB(1+C+C')+BC(1+A')+AC'(1+B')+A'C(1+B)\\\implies AB+BC+AC'+A'C$

$\implies$AB+A'C+AC'\