Changes

* The point <math>(1, 1, \ldots , 1, 1)</math> with all 1's as coordinates is the point where the conjunction of all posited variables evaluates to <math>1,\!</math> namely, the point where:

* The point <math>(1, 1, \ldots , 1, 1)</math> with all 1's as coordinates is the point where the conjunction of all posited variables evaluates to <math>1,\!</math> namely, the point where:

:: <math>x_1\ x_2\ \ldots\ x_{n-1}\ x_n = 1.</math>

:: <math>x_1 ~ x_2 ~\ldots~ x_{n-1} ~ x_n ~=~ 1.</math>

* The point <math>(0, 0, \ldots , 0, 0)</math> with all 0's as coordinates is the point where the conjunction of all negated variables evaluates to <math>1,\!</math> namely, the point where:

* The point <math>(0, 0, \ldots , 0, 0)</math> with all 0's as coordinates is the point where the conjunction of all negated variables evaluates to <math>1,\!</math> namely, the point where:

:: <math>(x_1)(x_2)\ldots(x_{n-1})(x_n) = 1.</math>

:: <math>\texttt{(} x_1 \texttt{)(} x_2 \texttt{)} \ldots \texttt{(} x_{n-1} \texttt{)(} x_n \texttt{)} ~=~ 1.</math>

To pass from these limiting examples to the general case, observe that a singular proposition <math>s : \mathbb{B}^k \to \mathbb{B}</math> can be given canonical expression as a conjunction of literals, <math>s = e_1 e_2 \ldots e_{k-1} e_k</math>.  Then the proposition <math>\nu (e_1, e_2, \ldots, e_{k-1}, e_k)</math> is <math>1\!</math> on the points adjacent to the point where <math>s\!</math> is <math>1,\!</math> and 0 everywhere else on the cube.

To pass from these limiting examples to the general case, observe that a singular proposition <math>s : \mathbb{B}^k \to \mathbb{B}</math> can be given canonical expression as a conjunction of literals, <math>s = e_1 e_2 \ldots e_{k-1} e_k</math>.  Then the proposition <math>\nu (e_1, e_2, \ldots, e_{k-1}, e_k)</math> is <math>1\!</math> on the points adjacent to the point where <math>s\!</math> is <math>1,\!</math> and 0 everywhere else on the cube.