A non-empty set B with two binary operations "+" and "." a unary operation " ' ", and two distinct element 0 and 1 is called a boolean algebra, denoted by (B, +, ., ', 0, 1) if and only if the following 4 properties are satisfied.
Axioms of boolean algebra-
If a,b,c ∈ B then
Let a, b and c ∈ B then
To prove the basic theorems we use the help of axioms.
proving a + a = a
R.H.S = a
= a + 0 by Identity laws
[ a + 0 = a]
= a + a.a' using complement law
[a.a' = 0]
= (a + a) . (a + a') using distributive laws
= (a + a) . 1 using complement laws
[a + a' = 1]
= (a + a) using Identity laws
[x . 1 = x, here x = (a + a)]
= L.H.S
Hence, a + a = a
proving a.a = a
L.H.S = a.a
= a.a + 0 by Identity laws
= a.a + a.a' using Complement laws
= a.(a + a') using distributive laws
= a.1 using Complement laws
= a using Identity laws
= R.H.S
Hence, a.a = a
proving a + 1 = 1
R.H.S = 1
= a + a' by Complement laws
= a + a'.1 using Identity laws
= (a + a').(a + 1) using Distributing laws
= 1.(a + 1) using Complement laws
= (a + 1).1 using commutative laws
= a + 1 using Identity laws
= L.H.S
Hence, a + 1 = 1
proving a.0 = 0
R.H.S = 0
= a.a' by Complement laws
= a.(a' + 0) using Identity laws
= (a.a') + (a.0) using Distributive laws
= 0 + (a.0) using Complement laws
= (a.0) + 0 using Commutative laws
= a.0 using Identity laws
= L.H.S
Hence, a.0 = 0
proving a+(a.b) = a
L.H.S = a+(a.b)
= a.1 + (a.b) by Identity laws
= a.(1 + b) using Distributive laws
= a.(b + 1) using Commutative laws
= a.1 using Boundedness laws
= a using Identity laws
= R.H.S
Hence, a+(a.b) = a
proving a.(a+b) = a
L.H.S = a.(a+b)
= a + 0 . (a+b) using Identity laws
= a+(0.b) using Distributive laws
= a using Boundedness laws
= L.H.S
Hence, a.(a+b) = a
proving (a+b)+c = a+(b+c)
R.H.S = a+(b+c)
= [a+(b+c)].1 by Identity law
= [a+(b+c)].(c+c') by Complement law
= {[a+(b+c)].c} + {[a+(b+c)].c'} by Distributive law
= {c.[a+(b+c)]} + {c'.[a+(b+c)]} by Commutative law
= [c.a+c.(b+c)] + [c'.a+c'(b+c)] by Distributive law
= (c.a + c) + (c'.a+c'(b+c)) using Commutive law and Absorption law
= c + (c'.a+c'(b+c)) using Commutive law and Absorption law
= c + (c'a + c'.b + c'.c) using Distributive law
= c + (c'.a + c'.b + c.c') using Commutative law
= c + (c'.a + c'.b + 0) using Complement law
= c + (c'.a + c'.b) using Identity law
= c + (c'.(a+b)) using Distributive law
= (c+c').(c+(a+b)) using Distributive law
= 1.[c+(a+b)] using Complement law
= [c+(a+b)].1 using Commutative law
= c + (a + b) using Identity law
= (a+b) + c using commutative law
= L.H.S
Hence (a+b)+c = a+(b+c)
proving (a.b).c = a.(b.c)
R.H.S = a.(b.c)
= [a.(b.c)]+0 using Identity law
= [a.(b.c)] + (c.c') by Complement law
= {[a.(b.c)]+c} . {[a.(b.c)]+c'} using Distributive law
= {c+[a.(b.c)]}.{c'+[a.(b.c)]} using Commutative law
= {(c+a).[c+(b.c)]}.{(c'+a).[c'a+(b.c)]} using Distributing law
= {(c+a).c}.{(c'+a).[c'+(b.c)]} using Commutative & Absorption law
= {c.(c+a)}.{(c'+a).[c'+(b.c)]} using Commutative law
= c . {(c'+a).[c'+(b.c)]} using Absorption law
= c . {(c'+a).[(c'+b).(c'+c)]} using distributive law
= c . {(c'+a).[(c'+b).1]} using Commutative and Complement laws
= c . {(c'+a).(c'+b)} using Identity law
= c . [c'+(a.b)] using Distributive law
= (c.c')+[c.(a.b)] using Distributive law
= 0 + [c.(a.b)] using Complement law
= c.(a.b) using Identity law
= (a.b).c using Commutative law
= L.H.S
Hence, (a.b).c = a.(b.c)
proving (a+b)' = a'.b'
we know a + a' = 1 and a.a' = 0
So, if we can prove that (a+b) + (a'.b') = 1 and (a+b).(a'.b') = 0 that means complement of (a+b) is a'.b'
First we will prove (a+b) + (a'.b') = 1
L.H.S = (a+b) + (a'.b')
= (a+b+a').(a+b+b') using Distributive law
= (b+a+a').(a+b+b') using Commutative law
= (b+1).(a+1) using Complement law
= 1.1 using Boundedness law
= 1 using Idemponent law
= R.H.S
Now we'll prove (a+b).(a'.b') = 0
L.H.S = (a+b).(a'.b')
= (a.a'.b') + (b.a'.b') by Distributive law from right side
= (a.a'.b') + (b.b'.a) using Commutative law
= (0.b') + (0.a) [as a.a' = 0]
= 0 + 0 [as a.0 = 0]
= 0 using Idemponent law
= R.H.S
Hence, as we have prove both the condition so we can say (a+b)' = a'.b'
proving (a.b)' = a' + b'
we know a + a' = 1 and a.a' = 0
So, if we can prove that (a.b) + (a' + b') = 1 and (a.b).(a' + b') = 0 that means complement of (a.b) is a'+b'
First we will prove (a.b) + (a'+b') = 1
L.H.S = (a.b) + (a'+b')
= (a'+b')+(a.b) by Commutative law
= (a'+b'+a).(a'+b'+b) by Distributive law
= (a'+a+b').(a'+b'+b) by Commutative law
= (1+b').(a'+1) by Complement law
= 1.1 by Boundedness law
= 1 by Idempodent law
= R.H.S
Now we will prove (a.b).(a'+b') = 0
L.H.S = (a.b).(a'+b')
= (a.b.a') + (a.b.b') by Distributive law
= (a.a'.b) + (a.b.b') by Commutative law
= (0.b) + (a.0) by Complement law
= 0 + 0 by Boundedness law
= 0 by Idempodent law
= R.H.S
Hence, as we have prove both the condition so we can say (a.b)' = a' + b'
1. (CB + CC).(A + B + C)
(C.B + C.C).(A + B + C)
= (CB + C).(A + B + C) [∵ C.C = C]
= C . (A + B + C) [∵ CB + C = C]
= C.A + C.B + C.C by distributive law
= C.A + C.B + C [∵ C.C = C]
= C.A + C [∵ C.B + C = C]
= C [∵ C.A + C = C]
xy + x'z + yz
xy + x'z + yz
= xy + x'z + yz(x+x') [∵ x+x' = 1]
= xy + x'z + xyz + x'yz
= xy + xyz + x'z + x'yz
= xy(1+z) + x'z(1+y)
= xy + x'z by Boundedness
A+B(A+B) + A(A'+B)
A+B(A+B) + A(A'+B)
= A + AB + BB + AA' + AB
= A + AB + B + 0 + AB [∵ BB = B & AA' = 0]
= A + AB + B [∵ X+X = X here X = AB]
= A + B [∵ A + AB = A]
Minterm of the two variables x & y are :
xy m3
x'y m1
xy' m2
x'y' m0
For three terms x, y & z [here x = 1 and x' = 0]
Product term minterm representation
x'y'z' m0
x'y'z m1
x'y z' m2
x'y z m3
x y'z' m4
x y'z m5
x y z' m6
x y z m7
Sum terms Maxterm representation [here x' = 1 and x = 0]
x + y + z M0
x + y + z' M1
x + y' + z M2
x + y' + z' M3
x' + y + z M4
x' + y + z' M5
x' + y' + z M6
x' + y' + z' M7
References ↓