For this answer I will be using the list of axioms provided by the following link
https://www.cs.tau.ac.il/~nin/Courses/ComStruct04/BooleanAlgebra.htm
Suppose there is a set S with two elements called x and y with two operations called + and * respectively.
Axiom #1 gives us closure of the operation which assures us that
1) Either x+y=x, or x+y=y (both can’t simultaneously be true: it’s one or the other)
2) Either x*y=x, or x*y=y (again one or the other is true but not both)
We needn’t worry about y+x or y*x since both operations are commutative by Axiom #3
Axiom #2 in the provided link is already used in the definition of S: it has two unique elements using the convention that we don’t list repeated elements in a set.
Axiom #5:
1) The operation + has an identity element called 0 such that for all elements x in S it follows that x+0=x
2) The operation * has an identity element called 1 such that for all elements x in S it follows that x*1=x
Since axiom 5 gives us two elements that S must have then these must be the only two elements that S contains.
S={0, 1}
Axiom #6:
Each element x in S must have a complement denoted by x’ such that
1) x+x’=1
2) x*x’=0
Let x=0, then we must have 0+0’=1; however, by axiom 5 we know that 0 is the additive identity so 0+0’=0’. Equating both sides of this equation we get 0’=1.
Therefore, 1 is the complement of 0.
Q.E.D
