PhyslibAlpha.ProbabilisticTheory.Effect.Complement

Given an effect $e$ in the order interval $[0, 1]$ of a partially ordered type $E$ with zero and unit elements, the complement of $e$ is the effect defined as $1 - e$. Physically, if $e$ represents the probability of a "yes" outcome in a test, its complement represents the probability of the "no" outcome.

Let $E$ be a partially ordered type with a zero element $0$ and a unit element $1$. For any effect $e$ in the order interval $[0, 1]$, the complement of the complement of $e$ is equal to $e$. That is, $\text{complement}(\text{complement}(e)) = e$. In physical terms, taking the "no" test of a "no" test returns the original "yes" test.

Let $E$ be an ordered additive commutative group with a unit element $1$. For any two effects $e_1, e_2 \in E$ (defined as elements satisfying $0 \le e \le 1$), the complement operation $e \mapsto 1 - e$ reverses the order. That is, if $e_1 \le e_2$, then $1 - e_2 \le 1 - e_1$.

For the set of effects defined on the order interval $[0, 1]$ in a partially ordered type $E$, the complement of the zero effect $0$ is the unit effect $1$.

In the context of probabilistic theories, let the set of effects be the order interval $[0, 1]$ in a partially ordered type $E$. The complement of the unit effect $1$ is the zero effect $0$.

Let $E$ be a partially ordered additive commutative group with a unit element $1$. For any effects $e, f \in E$ (where an effect is an element in the interval $[0, 1]$) and any $t$ in the unit interval $[0, 1]$, taking the complement commutes with the mixing operation. Specifically, the complement of the mixture of $e$ and $f$ with weight $t$ is equal to the mixture of the complements of $e$ and $f$ with the same weight $t$: $$1 - (t e + (1 - t) f) = t(1 - e) + (1 - t)(1 - f)$$