Surreal number/Simplicity

$\{2{\tfrac {1}{2}}|4{\tfrac {1}{2}}\}=3$, not ${\color {Red}3{\tfrac {1}{2}}}$ (which is the average of $2{\tfrac {1}{2}}$ and $4{\tfrac {1}{2}}$)

■ Surreal number/Simplicity illustrates that illustrates a case where the surreal number $x$ is not midway bwetten $X_{L}$ and $X_{R}$ (as it is for { 0 | 1 } = 1/2, { 0 | 1/2 } = 1/4, and { 1/2 | 1 } = 3/4.)

Instead, the simplicity rule yields:

$\{2{\tfrac {1}{2}}|4{\tfrac {1}{2}}\}=\{2|\;\}=3,$

which is the oldest number between $2{\tfrac {1}{2}}$ and $4{\tfrac {1}{2}}.$

...the rest is under construction...💧

Tøndering uses the convention that

x ≤ y ⇔ ¬∃ x_L ∈ X_L : y ≤ x_L ∧ ¬∃ y_R ∈ Y_R : y_R ≤ x,

where X_L and Y_R are the left set of x and the right set of y, respectively.

More scratchwork taken from Tøndering :

Notational convention. For two sets of surreal numbers, A and B, and a surreal number, c, we define the relations <, ≤, >, ≥, ≮, , ≯, and thus: A ≤ c if and only if ∀ a ∈ A : a ≤ c, c ≤ A if and only if ∀ a ∈ A : c ≤ a, A ≤ B if and only if ∀ a ∈ A ∀ b ∈ B : a ≤ b, and similarly for the other relations. Example. {1, 3, 5} < {6, 7} because both 1, 3, and 5 are less than 6 and 7. {3, 5, 6} ≮ 1 because neither 3, 5, and 6 is less than 1. Note that when sets are involved, ¬(A ≤ b) is not equivalent to A b. For example, neither {3, 5} ≤ 4 nor {3, 5} 4 is true. It follows from the definition above that ∅ ≤ b is always true, regardless of the value of b; and so is ∅ ≰ b, ∅ > b, etc.