# Thread: A Christmas tree for socionics

1. ## A Christmas tree for socionics

Hi all!

I designed a "Christmas tree" in order to help you to understand the bonds between intertype relationships. Concretely, knowing your intertype relationship with two people you have already typed, which intertype relationship these people are gonna experience?
christmasTree.png

To keep the image readable I represent each symmetric intertype relationship (asymmetric ones are much trickier to represent) with one character, with a legend similar to Ibrahim Tencer's article:A = activation;

• C = conflict;
• D = duality;
• G = super-ego;
• H = half-duality;
• I = illusion or mirage;
• K = kinship;
• L = look-a-like or business;
• M = mirror;
• Q = quasi-identical;
• X = extinguishment.

E is Identity, that does not change any intertype relationship it is combined with. Thus it is represented alone in the "ground".

As you can see there are two kinds of lines. Continuous lines (or ellipses in the center) tell you that if you meet two intertype relationships then the third in the line is the result of the meeting of both intertype relationships. For example I – G give H, meaning that if your mirage meets your super-ego they will experience half-duality. Similarly, C – D – Q line means the dual of your conflictor is your quasi-identical. In more mathematical terms continuous lines tell you how to multiply symmetric intertype relationships.

How about the dashed lines, oriented with an arrow named 'B'? These lines represent benefit rings. When considering two neighbouring intertype relationships on a benefit ring, benefactor comes first and beneficiary comes second, e.g. A is benefactor of K.

Only two benefit rings are represented such: A > K > Q > L > A and M > H > C > I > M. The two other benefit rings, which are not represented, involve asymmetric relationships: E > b > G > B > E and D > s > X > S > D where B/b and S/s are respectively Benefactor/beneficiary and Supervisor/supervisee.

Therefore you can deduce that the conflictor is the beneficiary of the half-dual or that the beneficiary of the quasi-identical is the look-a-like.

Now let us go to more headache with supervision relationships.
If two intertype relationships are not bound by any kind of line, then they are in Supervision. To calculate who supervises whom, one should use the fact that supervisor is dual of benefactor, and supervisee is dual of beneficiary. Hence, if two relationships are not bound by any kind of line, continuous or dashed, then take the dual of one and see where it ends.

Example 1: relationship between your activator A and your half-dual H. Consider dual of A : A – D – M gives you mirror. Then look at M's position relative to H: your mirror is benefactor of your half-dual, hence your activator is the supervisor of your half-dual.

How about doing it in the other sense, aka dual of H? Following dual line H – D – K gives K as dual of H. But K is beneficiary of A, hence same conclusion as above.

Example 2: the supervisee of your quasi-identical is the dual of the beneficiary of your quasi-identical : Q goes to L through benefit ring, then follow L – D line leads you to I: the supervisee of your quasi-identical is therefore your mirage.

Still here?

Serious Christmas

2. very nice.

3. I popped a fuse at the Benefit rings. Merry Whatever!

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