Was Lewis Carroll an Amazing Oppositional Geometer? (2014), p. 403
by Moretti, Alessio
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Caption
- Each oppositional tetrahexahedron has an 'envelope' made of 'independence relations'.
- Aristotelian family
- Classical Sigma-7
- 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
- Rhombic Dodecahedron (irregular)
- Colinearity range
- 0
- Coplanarity range
- 0
- Cospatiality range
- 0
- Representation of contradiction
- By central symmetry
Logic
Geometry
- 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
Vertex description
Edge description
- Diagram is colored
- Yes
- Diagram is embellished
- No
- Tags
- Boolean closed ;
- 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 indicial notation) can be represented as follows:
1000 = $ x_1y'_0 $
0100 = $ xy_1\ \dag \ xy'_1 $
0010 = $ x_1y_0 $
0001 = $ x_0 $
1100 = $ xy_1 $
1010 = $ x_1y_0\ § \ x_1y'_0 $
1001 = $ xy'_0 $
0110 = $ xy'_1 $
0101 = $ x_0\ § \ (xy_1\ \dag \ xy'_1) $
0011 = $ xy_0 $
1110 = $ x_1 $
1101 = $ x_0y_1 $
1011 = $ xy_0 \ § \ xy'_0 $
0111 = $ x_0y'_1 $
Note: $x'$ means not-$x$, $\dag$ means "and", $§$ means "or" (cf. footnote 5 on p. 387).