Proof of commutative property in Boolean algebra a∨b = b∨a, a∧b = b∧a, Does anybody know how to prove commutative property in Boolean algebra?
$a \vee b= b \vee a$
$a \wedge b = b \wedge a$
Thank you, have a nice day.
Edit:
Algebra $(A,\wedge,\vee,',0,1)$ 
$\vee, \wedge$ are binary operations, $'$ is unary operation, $0,1$ are constants


*

*Associativity of binary operations: $(a \wedge b) \wedge c = a \wedge (b \wedge c)$, $(a \vee b) \vee c = a \vee (b \vee c)$

*Distributivity: $a\wedge(b\vee c) = (a\wedge b)\vee(a\wedge c)$, $(b\vee c)\wedge a = (b\vee a)\wedge (c \vee a)$

*$'$: $(a')\wedge a = a\wedge(a') = 0$, $(a')\vee a = a\vee (a') = 1$

*$0$: $a\wedge 0 = 0\wedge a = 0$, $0\vee a = a\vee 0 = a$

*$1$: $a\wedge 1 = 1\wedge a = a$, $1\vee a = a\vee 1 = 1$

 A: First we prove idempotency ($a=a\lor a$), though we might not need it later on.
$$a=a\lor 0=a\lor(a\land a')=(a\lor a)\land (a\lor a')=(a\lor a)\land 1=a\lor a$$
Second, we prove uniqueness of the complement, in the sense that 
$a\land b=0,\ b\lor a=1\implies b=a'$:
$$a'=a'\land 1=a'\land (b\lor a)=(a'\land b)\lor (a'\land a)=a'\land b\\
b=1\land b=(a\lor a')\land b= (a\land b)\lor (a'\land b)=a'\land b $$
In particular, it implies $a''=a$.
Then certain forms of absorbance follows: $a=a\lor(b\land a)$ 
$$a'\lor (a\lor(b\land a))=1\\
(a\lor(b\land a))\land a'=(a\land a')\lor(b\land a\land a')=0 $$
We similarly get $a=(a\land b)\lor a$, and two other equations by duality.
Then, we get a key lemma: $a\lor b=1 \implies a'=a'\land b \implies b\lor a=1$: 
Supposed $a\lor b=1$, we get $a'=a'\land 1=a'\land(a\lor b)=(a'\land a)\lor (a'\land b)=a'\land b$. 
Supposed $a'=a'\land b$, we get $b\lor a'=b\lor (a'\land b)=b$ by absorbance, so $\ b\lor a=(b\lor a')\lor a=1$.
Note that this implies $a\lor x\lor a'=1$ for any $x$, as we have $\ (x\lor a')\lor a=1$. 
Using their dual statements as well ($a\land b=0\iff b\land a=0$ and $a'\land x\land a=0$), we can finally arrive to commutativity by observing that both $a\lor b$ and $b\lor a$ are complements of $a'\land b'$:
$$(a\lor b)\lor (a'\land b')=(a\lor b\lor a')\land (a\lor b\lor b')=1\\
(b\lor a)\lor (a'\land b')=(b\lor a\lor a')\land (b\lor a\lor b')=1\\
-\cdot-\cdot- \\
(a'\land b')\land (a\lor b)=(a'\land b'\land a)\lor(a'\land b'\land b)=0\\
(a'\land b')\land (b\lor a)=(a'\land b'\land b)\lor(a'\land b'\land a)=0
$$
