3-way transistor and uneven logic

Lets think about this: a 3 way transistor and uneven logic.

So we have, logically speaking:

a&b&c with three possible values (0,1,2):

0&0&0=0

0&0&1=0

0&0&2=0

0&1&0=0

0&1&1=1

0&1&2=0

0&2&0=0

0&2&1=0

0&2&2=2

1&0&0=0

1&0&1=1

1&0&2=0

1&1&0=1

1&1&1=1

1&1&2=1

1&2&0=0

1&2&1=1

1&2&2=2

2&0&0=0

2&0&1=0

2&0&2=2

2&1&0=0

2&1&1=1

2&1&2=2

2&2&0=2

2&2&1=2

2&2&2=2

Right?

Let's update with the rest:

Updating on the 4 bit propositional logic,

Which should be 3 way or 2nd order uneven logic

(since 1=1 is 1st order):

As in: What if there were 3 true/false/X values?

Now for or denoted by | (inclusive):

0|0|0=0

0|0|1=1

0|1|0=1

0|1|1=1

1|0|0=1

1|0|1=1

1|0|2=2

1|1|0=1

1|1|1=1

1|1|2=1

-

0|0|2=2

0|1|2=2

0|2|0=2

0|2|1=2

2|0|0=2

2|0|1=2

2|1|0=2

2|1|1=2

2|1|2=2

2|2|2=2

Let's denote exclusive or with /:

0/0/0=0

0/0/1=1

0/0/2=2

0/1/0=1

0/1/1=0

0/1/2=2

1/0/0=1

1/0/1=0

1/0/2=0

1/1/0=0

1/1/1=0

1/1/2=0

1/2/0=0

1/2/1=0

1/2/2=0

2/0/0=2

2/0/1=0

2/1/0=0

2/2/0=0

2/2/1=0

2/2/2=0

--

Now for not denoted by !:

!0=1|2

!1=0|2

!2=0|1

Now, if then denoted by >:

0,0,0>0

0,0,1>0

0,0,2>0

0,1,0>0

0,1,1>0

0,1,2>0

1,0,0>0

1,0,1>0

1,0,2>0

1,1,0>0

1,1,1>1

1,1,2>0

1,2,0>0

1,2,1>0

1,2,2>0

2,0,0>0

2,0,1>0

2,1,0>0

2,1,1>0

2,1,2>0

2,2,0>0

2,2,1>0

2,2,2>2

Now,

To be precise we need to have a 3-way operator for & denoting ALL THREE VALUES ARE TRUE.

Let's for now use # to denote it (behaving like & or AND in normal logic):

Basically every option equates to 0, except:

1#1#1=1

2#2#2=2