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User:Jon Awbrey/Cactus Language Stretching Exercises (view source)
Revision as of 14:18, 4 November 2025
, Yesterday at 14:18→Cactus Language • Stretching Exercises: adjust line spacing [4pt]
[| \downharpoonleft s \downharpoonright |]
[| \downharpoonleft s \downharpoonright |]
& = & [| F |]
& = & [| F |]
\\[6pt]
\\[4pt]
& = & F^{-1} (1)
& = & F^{-1} (1)
\\[6pt]
\\[4pt]
& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ s ~\}
& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ s ~\}
\\[6pt]
\\[4pt]
& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ F(x, y) = 1 ~\}
& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ F(x, y) = 1 ~\}
\\[6pt]
\\[4pt]
& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ F(x, y) ~\}
& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ F(x, y) ~\}
\\[6pt]
\\[4pt]
& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ \texttt{(}~x~,~y~\texttt{)} = 1 ~\}
& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ \texttt{(}~x~,~y~\texttt{)} = 1 ~\}
\\[6pt]
\\[4pt]
& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ \texttt{(}~x~,~y~\texttt{)} ~\}
& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ \texttt{(}~x~,~y~\texttt{)} ~\}
\\[6pt]
\\[4pt]
& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ x ~\mathrm{exclusive~or}~ y ~\}
& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ x ~\mathrm{exclusive~or}~ y ~\}
\\[6pt]
\\[4pt]
& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ \mathrm{just~one~true~of}~ x, y ~\}
& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ \mathrm{just~one~true~of}~ x, y ~\}
\\[6pt]
\\[4pt]
& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ x ~\mathrm{not~equal~to}~ y ~\}
& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ x ~\mathrm{not~equal~to}~ y ~\}
\\[6pt]
\\[4pt]
& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ x \nLeftrightarrow y ~\}
& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ x \nLeftrightarrow y ~\}
\\[6pt]
\\[4pt]
& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ x \neq y ~\}
& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ x \neq y ~\}
\\[6pt]
\\[4pt]
& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ x + y ~\}.
& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ x + y ~\}.
\end{array}</math>
\end{array}</math>
|
|
<math>\begin{matrix}
<math>\begin{matrix}
p
p & = & \upharpoonleft P \upharpoonright & : & X \to \mathbb{B}
& = &
\\[4pt]
\upharpoonleft P \upharpoonright
q & = & \upharpoonleft Q \upharpoonright & : & X \to \mathbb{B}
& : &
\\[4pt]
X \to \mathbb{B}
(p, q) & = & (\upharpoonleft P \upharpoonright, \upharpoonleft Q \upharpoonright) & : & (X \to \mathbb{B})^2
\\
q
& = &
\upharpoonleft Q \upharpoonright
& : &
X \to \mathbb{B}
\\
(p, q)
& = &
(\upharpoonleft P \upharpoonright, \upharpoonleft Q \upharpoonright)
& : &
(X \to \mathbb{B})^2
\end{matrix}</math>
\end{matrix}</math>
|}
|}
|
|
<math>\begin{array}{ccccl}
<math>\begin{array}{ccccl}
F^\$
F^\$ & = & \underline{(} \ldots, \ldots \underline{)}^\$ & : & (X \to \mathbb{B})^2 \to (X \to \mathbb{B})
& = &
\\[4pt]
\underline{(} \ldots, \ldots \underline{)}^\$
F^\$ (p, q) & = & \underline{(}~p~,~q~\underline{)}^\$ & : & X \to \mathbb{B}
& : &
(X \to \mathbb{B})^2 \to (X \to \mathbb{B})
\\
F^\$ (p, q)
& = &
\underline{(}~p~,~q~\underline{)}^\$
& : &
X \to \mathbb{B}
\end{array}</math>
\end{array}</math>
|}
|}
<math>\begin{matrix}
<math>\begin{matrix}
F^\$ (p, q)(x) & = & \underline{(}~p~,~q~\underline{)}^\$ (x) & \in & \mathbb{B}
F^\$ (p, q)(x) & = & \underline{(}~p~,~q~\underline{)}^\$ (x) & \in & \mathbb{B}
\\
\\[4pt]
\Updownarrow & & \Updownarrow
\Updownarrow & & \Updownarrow
\\
\\[4pt]
F(p(x), q(x)) & = & \underline{(}~p(x)~,~q(x)~\underline{)} & \in & \mathbb{B}
F(p(x), q(x)) & = & \underline{(}~p(x)~,~q(x)~\underline{)} & \in & \mathbb{B}
\end{matrix}</math>
\end{matrix}</math>
|}
|}
[| F^\$ (p, q) |]
[| F^\$ (p, q) |]
& = & [| \underline{(}~p~,~q~\underline{)}^\$ |]
& = & [| \underline{(}~p~,~q~\underline{)}^\$ |]
\\[6pt]
\\[4pt]
& = & (F^\$ (p, q))^{-1} (1)
& = & (F^\$ (p, q))^{-1} (1)
\\[6pt]
\\[4pt]
& = & \{~ x \in X ~:~ F^\$ (p, q)(x) ~\}
& = & \{~ x \in X ~:~ F^\$ (p, q)(x) ~\}
\\[6pt]
\\[4pt]
& = & \{~ x \in X ~:~ \underline{(}~p~,~q~\underline{)}^\$ (x) ~\}
& = & \{~ x \in X ~:~ \underline{(}~p~,~q~\underline{)}^\$ (x) ~\}
\\[6pt]
\\[4pt]
& = & \{~ x \in X ~:~ \underline{(}~p(x)~,~q(x)~\underline{)} ~\}
& = & \{~ x \in X ~:~ \underline{(}~p(x)~,~q(x)~\underline{)} ~\}
\\[6pt]
\\[4pt]
& = & \{~ x \in X ~:~ p(x) + q(x) ~\}
& = & \{~ x \in X ~:~ p(x) + q(x) ~\}
\\[6pt]
\\[4pt]
& = & \{~ x \in X ~:~ p(x) \neq q(x) ~\}
& = & \{~ x \in X ~:~ p(x) \neq q(x) ~\}
\\[6pt]
\\[4pt]
& = & \{~ x \in X ~:~ \upharpoonleft P \upharpoonright (x) ~\neq~ \upharpoonleft Q \upharpoonright (x) ~\}
& = & \{~ x \in X ~:~ \upharpoonleft P \upharpoonright (x) ~\neq~ \upharpoonleft Q \upharpoonright (x) ~\}
\\[6pt]
\\[4pt]
& = & \{~ x \in X ~:~ x \in P ~\nLeftrightarrow~ x \in Q ~\}
& = & \{~ x \in X ~:~ x \in P ~\nLeftrightarrow~ x \in Q ~\}
\\[6pt]
\\[4pt]
& = & \{~ x \in X ~:~ x \in P\!-\!Q ~\mathrm{or}~ x \in Q\!-\!P ~\}
& = & \{~ x \in X ~:~ x \in P\!-\!Q ~\mathrm{or}~ x \in Q\!-\!P ~\}
\\[6pt]
\\[4pt]
& = & \{~ x \in X ~:~ x \in P\!-\!Q ~\cup~ Q\!-\!P ~\}
& = & \{~ x \in X ~:~ x \in P\!-\!Q ~\cup~ Q\!-\!P ~\}
\\[6pt]
\\[4pt]
& = & \{~ x \in X ~:~ x \in P + Q ~\}
& = & \{~ x \in X ~:~ x \in P + Q ~\}
\\[6pt]
\\[4pt]
& = & P + Q ~\subseteq~ X
& = & P + Q ~\subseteq~ X
\\[6pt]
\\[4pt]
& = & [|p|] + [|q|] ~\subseteq~ X
& = & [|p|] + [|q|] ~\subseteq~ X
\end{array}</math>
\end{array}</math>
|}
|}