# Was Lewis Carroll an Amazing Oppositional Geometer? (2014), p. 403

by Moretti, Alessio

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### Caption

- The twelve orange segments of independence let emerge three squares of a new kind.

- Aristotelian family
- Degenerate Sigma-2 with Unconnectedness 4
- 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
- Square (regular)
- 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
- subdiagram ;
- 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).