DHARM
N-COM\APPENDIX.PM5BOOLEAN ALGEBRA 355
Since, postulates are the basic axioms of algebraic structures and so they required no
proof. We can prove the listed theorems from the given postulates.
(T1)LHS
x + x = (x + x). 1 ∴ x. 1 = x P1
= (x + x). (x + x′) ∴ x + x′ = 1 P2
= x + x. x′∴ (x + y) (x + z) = x + y. z P4
= x + 0 ∴ x. x′ = 0 P2
= x ∴ x + 0 = x P1
RHS Hence proved.
Dual part of Theorem T1 can be proved similarly such as,
(T1′)LHS
x. x = x. x + 0 ∴ x + 0 = x P1
= x. x + x. x′∴ x. x′ = 0 P2
= x. (x + x′) ∴ x. y + x. z = x. (y + z) P4
= x. 1 ∴ x + x′ = 1 P2
= x ∴ x. 1 = x P1
RHS Hence, proved.
(T2) LHS
x + 1 = (x + 1). 1 ∴ x .1 = x P1
= (x + 1). (x + x′) ∴ x + x′ = 1 P2
= x + 1. x′∴ (x + y). (x + z) = x + y. z P4
= x + x′∴ 1. x′ = x′ P1
= 1 ∴ x + x′ = x P2
RHS Hence, proved.
(T2′)LHS
x. 0 = (x. 0) + 0 ∴ x + 0 = x P1
= (x. 0) + (x. x′) ∴ x. x′ = 0 P2
= x. (x′ + 0) ∴ x. y + x. z = x. (y + z) P4
= x. x′∴ x′ + 0 = x′ P1
= 0 ∴ x. x′ = 0 P2
RHS Hence, proved
(T3) Since, x + x′ = 1 and x. x′ = 0 (i) P2
Define the complement of x i.e. x′. So to determine the complement of x′ we have,
(x′)′ + x′ = 1 and (x′)′. x′ = 0 (ii) P2
On comparing (i) and (ii) we obtain,
(x′)′ = x
Hence, proved.