Relevance and irredundance

Consider the function $\varphi _1$ defined in Tab. 1 whose PDNF is

$$\varphi_1(\boldsymbol{z})=(z_2\wedge z_3)\vee (z_2\wedge z_3\wedge z_4)\vee (z_1\wedge z_3)\vee$$ (3)

$$(z_1\wedge z_3\wedge z_4)\vee (z_1\wedge z_2\wedge z_3)\vee (z_1\wedge z_2\wedge z_3\wedge z_4). \nonumber$$

By calculating the influence of the variables $z_1, \dots , z_4$ we obtain that

$$I_1(\varphi_1)=1/4,$$     $$I_2(\varphi_1)=1/4,$$     $$I_3(\varphi_1)=3/4,$$     $$I_4(\varphi_1)=0$$     $$$$

Table 1: Truth table of the positive Boolean function $\varphi _1$.

$\boldsymbol{z}$				$\varphi_1(\boldsymbol{z})$	$\boldsymbol{z}$				$\varphi_1(\boldsymbol{z})$
$z_1$	$z_2$	$z_3$	$z_4$		$z_1$	$z_2$	$z_3$	$z_4$
0	0	0	0	0	$1$	0	0	0	0
0	0	0	$1$	0	$1$	0	0	$1$	0
0	0	$1$	0	0	$1$	0	$1$	0	$1$
0	0	$1$	$1$	0	$1$	0	$1$	$1$	$1$
0	$1$	0	0	0	$1$	$1$	0	0	0
0	$1$	0	$1$	0	$1$	$1$	0	$1$	0
0	$1$	$1$	0	$1$	$1$	$1$	$1$	0	$1$
0	$1$	$1$	$1$	$1$	$1$	$1$	$1$	$1$	$1$

i.e. $z_1,z_2$ and $z_3$ are relevant, while $z_4$ is irrelevant.

Then, considering that every positive Boolean function can be expressed only through its relevant variable, we obtain that $\varphi _1$ can be equivalently described by using only three variables (Tab. 2). Its irredundant PDNF is

$$\varphi_1(\boldsymbol{z})=(z_2\wedge z_3)\vee (z_1\wedge z_3)$$ (4)

Furthermore, the following expression profile, including $z_1$, $z_2$ and $z_3$, gives a compact representation of $\varphi _1$.

$$\varphi_1(\boldsymbol{z})=[(z_1,z_3)_2,(z_2,z_3)_2]_1$$ (5)

Table 2: Truth table of the function $\varphi _1$ composed only by relevant variables

$\boldsymbol{z}$			$\varphi_1(\boldsymbol{z})$
$z_1$	$z_2$	$z_3$
0	0	0	0
0	0	$1$	0
0	$1$	0	0
0	$1$	$1$	$1$
$1$	0	0	0
$1$	0	$1$	$1$
$1$	$1$	0	0
$1$	$1$	$1$	$1$