You're using an ancient browser to surf the modern web. Please update to the latest version (and don't use Internet Explorer!).

Leonardi.DB
a logical geometry project

A Cube of Opposition for Predicate Logic (2020), p. 105
by Nilsson, Jørgen Fischer

Caption

Cube for relational predicate logic

Logic

Aristotelian family
Buridan Sigma-4
Boolean complexity
6
Number of labels per vertex (at most)
2
Equivalence between (some) labels of the same vertex
Yes
Analogy between (some) labels of the same vertex
No
Uniqueness of the vertices up to logical equivalence
Yes
Errors in the diagram
No

Geometry

Shape
Cube (regular)
Colinearity range
0
Coplanarity range
0
Cospatiality range
0
Representation of contradiction
By some other geometric feature

Vertex description

Conceptual info
No
Mnemonic support (AEIO, purpurea ...)
No
Form
none
Label type
symbolic
Symbolic field
logic
Contains partial formulas or symbols
Yes
Logical system
predicate logic

Edge description

Style

Diagram is colored
No
Diagram is embellished
No
Tags
composed operator duality

Additional notes

Let $C,D$ be unary predicates and $R$ a binary relation.

$\forall\forall$ stands for $\forall x( Cx \to \forall y (Dy \to Rxy)$
$\forall\exists$ stands for $\forall x( Cx \to \exists y (Dy \wedge Rxy)$
$\exists\forall$ stands for $\exists x( Cx \wedge \forall y (Dy \to Rxy)$
$\exists\exists$ stands for $\exists x( Cx \wedge \exists y (Dy \wedge Rxy)$

$\forall\forall\neg$ stands for $\forall x( Cx \to \forall y (Dy \to \neg Rxy)$
$\neg\exists\exists$ stands for $\neg\exists x( Cx \wedge \exists y (Dy \wedge Rxy)$

$\forall\exists\neg$ stands for $\forall x( Cx \to \exists y (Dy \wedge \neg Rxy)$
$\neg\exists\forall$ stands for $\neg\exists x( Cx \wedge \forall y (Dy \to Rxy)$

$\exists\forall\neg$ stands for $\exists x( Cx \wedge \forall y (Dy \to\neg Rxy)$
$\neg\forall\exists$ stands for $\neg\forall x( Cx \to \exists y (Dy \wedge Rxy)$

$\exists\exists\neg$ stands for $\exists x( Cx \wedge \exists y (Dy \wedge\neg Rxy)$
$\neg\forall\forall$ stands for $\neg\forall x( Cx \to \forall y (Dy \to Rxy)$

(Cf. pp. 104-105.)
↑ Back to top ↑