ThmDex – An index of mathematical definitions, results, and conjectures.
Result R4308 on D527: Composite map
Map composition need not be commutative
Formulation 0
Let $f, g : \{ 0, 1 \} \to \{ 0, 1 \}$ each be a D18: Map such that
(i) \begin{equation} f(0) = f(1) = 0 \end{equation}
(ii) \begin{equation} g(0) = g(1) = 1 \end{equation}
Then
(1) \begin{equation} g(f(0)) = g(f(1)) = g(0) = 1 \end{equation}
(2) \begin{equation} f(g(0)) = f(g(1)) = f(1) = 0 \end{equation}
Proofs
Proof 0
Let $f, g : \{ 0, 1 \} \to \{ 0, 1 \}$ each be a D18: Map such that
(i) \begin{equation} f(0) = f(1) = 0 \end{equation}
(ii) \begin{equation} g(0) = g(1) = 1 \end{equation}
Clear. $\square$