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Leonardi.DB
a logical geometry project

Was Lewis Carroll an Amazing Oppositional Geometer? (2014), p. 394 by Moretti, Alessio

Caption

Each of the six vertices lying in the middle of one of the edges has its contradictory negation in the vertex lying in the middle of the edge opposed to it.

Logic

Aristotelian family
Degenerate Sigma-3 with Unconnectedness 12
Boolean complexity
4
Number of labels per vertex (at most)
1
Uniqueness of the vertices up to logical equivalence
Yes
Errors in the diagram
No

Geometry

Shape
Octahedron (regular)
Colinearity range
0
Coplanarity range
0
Cospatiality range
0
By central symmetry

Vertex description

Conceptual info
No
Mnemonic support (AEIO, purpurea ...)
No
Form
dots
Label type
symbolic
Symbolic field
logic
Contains partial formulas or symbols
Yes
Logical system
predicate logic

Style

Diagram is colored
No
Diagram is embellished
No
Tags
subdiagram
;
existential import

Consider the following partition:
1) A! (all x are y, and there is at least one x)
2) I $\wedge$ O (some x are y and some x are not y)
3) E! (no x are y, and there is at least one x)
4) there are no x

With the bitstrings based on this partition, the formulas of this diagram (in Greek notation) can be represented as follows:

1000 = $\alpha\beta\delta$
0100 = $\alpha\beta\gamma$
0010 = $\alpha\gamma\delta$
0001 = $\beta\gamma\delta$
1100 = $\alpha\beta$
1010 = $\alpha\delta$
1001 = $\beta\delta$
0110 = $\alpha\gamma$
0101 = $\beta\gamma$
0011 = $\gamma\delta$
1110 = $\alpha$
1101 = $\beta$
1011 = $\delta$
0111 = $\gamma$

Note: $\alpha\beta$ means $\alpha\wedge\beta$ (cf. p. 391).

Note: the octahedron shape is easiest to see if the encompassing tetrahedron is tilted so that it stands on one of its edges.