Answer:
Simplification of the expressions:
a) a’b’ + ab + ab' = a +b'
b) r’ + rt + rs’ + rs’t’ = r' + s' + t
c) (x + y)’(x’ + y’)’ = False
d) b'+ bc + ab + a'bc' = True
e) ac + b'd + a'bc + b'cd' = c + b'd
Step-by-step explanation:
The step by step solution for each expression will use the following laws of Boolean Algebra:
Idempotent Law:
aa=a
a+a=a
Associative Law
:
(ab)c=a(bc)
(a+b)+c=a+(b+c)
Commutative Law
:
ab=ba
a+b=b+a
Distributive Law
:
a(b+c)=ab+ac
a+(bc)=(a+b)(a+c)
Identity Law
:
a*0=0 a*1=a
a+1=1 a+0=a
Complement Law
:
aa'=0
a+a'=1
Involution Law
:
(a')'=a
DeMorgan's Law
:
(ab)'=a'+b'
(a+b)'=a'b'
Absorption Law:
a+(ab)=a
a(a+b)=a
(ab)+(ab')=a
(a+b)(a+b')=a
a+(a'b)=a+b
a(a'+b)a*b
Step by step Solution:
a) F(a,b) = a’b’ + ab + ab'
a(b+b')+a'b' Commutative Law
a+a'b Complement Law
F(a,b)=a+b' Absorption Law
b) F(r,s,t) = r’ + rt + rs’ + rs’t’
(r'+rs')+rt+rs't' Absorption Law
r'+s'+rt+rs't' Distributive Law
r'+s'+rt+s' Absorption Law
r'+s'+rt Absorption Law
F(r,s,t) = r'+s'+t Absorption Law
c) F(x,y) = (x + y)’(x’ + y’)’
(x'y')(x''y'') DeMorgan's Law
(x'y')xy Involution Law
x'(y'x)y Associative Law
x'(xy')y Commutative Law
(x'x)(y'y) Associative Law
(0)(0) Complement Law
F(x,y)=False
d) F(a,b,c) = b'+ bc + ab + a'bc'
b'+c+b(a+a'c') Absorption Law
b'+c+b(a+c') Absorption Law
b'+c+ba+bc' Distributive Law
(b'+ba)+(c+bc') Associative Law
b'+a+c+b Absorption Law
1+a+c Complement Law
F(a,b,c)=True
e) F(a,b,c,d) = ac + b'd + a'bc + b'cd'
ac+a'bc+b'd+b'cd' Commutative Law
c(a+a'b)+b'(d+cd') Associative and Distributive Law
c(a+b)+b'(d+c) Absorption Law
ac+bc+b'd+b'c Distributive Law
ac+(bc+b'c)+b'd Associative and Commutative Law
ac+c(b+b')+b'd Associative and Distributive Law
ac+c*1+b'd Complement Law
c(a+1)+b'd Distributive and Identity Law
F(a,b,c,d)=c+b'd