Was Lewis Carroll an Amazing Oppositional Geometer? (2014), p. 394
by Moretti, Alessio
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Caption
- Each of the four exterior vertices $\alpha$, $\beta$, $\gamma$ and $\delta$ has its contradictory negation in the middle point of the opposite face.
- Aristotelian family
- Non-Sigma
- 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
- Shape
- Tetrahedron (regular)
- Colinearity range
- 0
- Coplanarity range
- 0–1
- Cospatiality range
- 0
- Representation of contradiction
- N.A.
Logic
Geometry
- Conceptual info
- No
- Mnemonic support (AEIO, purpurea ...)
- No
- Form
- dots ,
- none
- Label type
- symbolic ,
- none
- Symbolic field
- logic
- Contains partial formulas or symbols
- Yes
- Logical system
- predicate logic
Vertex description
Edge description
- Diagram is colored
- Yes
- Diagram is embellished
- No
- Tags
- existential import
Style
Additional notes
- 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).