Arabic numeral arithmetic: examples

It’s example time.

Example 1

For the first example let’s show that $⌜537⌝$ is an $np$ (and that it has a particular interpretation in $\mathcal{M}_{ANA}$).

\[\begin{prooftree} \AxiomC{$⟨⌜5⌝, I_{ANA}(⌜5⌝)⟩ ⊢ atom$} \RightLabel{Rule 2}\UnaryInfC{$⟨⌜5⌝, I_{ANA}(⌜5⌝)⟩ ⊢ n$} \AxiomC{$⟨⌜3⌝, I_{ANA}(⌜3⌝)⟩ ⊢ atom$} \RightLabel{Rule 2}\UnaryInfC{$⟨⌜3⌝, I_{ANA}(⌜3⌝)⟩ ⊢ n$} \AxiomC{$⟨⌜7⌝, I_{ANA}(⌜7⌝)⟩ ⊢ atom$} \RightLabel{Rule 2}\UnaryInfC{$⟨⌜7⌝, I_{ANA}(⌜7⌝)⟩ ⊢ n$} \RightLabel{Rule 3}\BinaryInfC{$⟨⌜37⌝, I_{ANA}(⌜3⌝) * 10^{length(⌜7⌝)} + I_{ANA}(⌜7⌝)⟩ ⊢ n$} \RightLabel{Rule 3}\BinaryInfC{$⟨⌜537⌝, I_{ANA}(⌜5⌝) * 10^{length(⌜37⌝)} + I_{ANA}(⌜3⌝) * 10^{length(⌜7⌝)} + I_{ANA}(⌜7⌝)⟩ ⊢ n$} \RightLabel{Rule 4}\UnaryInfC{$⟨⌜537⌝, I_{ANA}(⌜5⌝) * 10^{length(⌜37⌝)} + I_{ANA}(⌜3⌝) * 10^{length(⌜7⌝)} + I_{ANA}(⌜7⌝)⟩ ⊢ np$} \end{prooftree}\]

What’s proved is the statement in the very last line of this derivation: that the string $⌜537⌝$ has the interpretation \[I_{ANA}(⌜5⌝) * 10^{length(⌜37⌝)} + I_{ANA}(⌜3⌝) * 10^{length(⌜7⌝)} + I_{ANA}(⌜7⌝)\] and the category $np$. Since we know that

$I_{ANA}(⌜5⌝) = 5$,
$I_{ANA}(⌜3⌝) = 3$, and
$I_{ANA}(⌜7⌝) = 7$

as well as what $length(⌜37⌝)$ and $length(⌜7⌝)$ are—$2$ and $1$, respectively—the very last line of this proof/derivation can be rewritten as

\[5 * 10^{2} + 3 * 10^{1} + 7\]

which, of course, is $537$.

Example 2

For the second example, we’ll show that $⌜((37 + 1) × 0) = 5⌝$ is an $s$, and also what its interpretation in $\mathcal{M}_{ANA}$ is.

\[\begin{prooftree} \AxiomC{$⟨⌜3⌝, I_{ANA}(⌜3⌝)⟩ ⊢ atom$} \RightLabel{Rule 2}\UnaryInfC{$⟨⌜3⌝, I_{ANA}(⌜3⌝)⟩ ⊢ n$} \AxiomC{$⟨⌜7⌝, I_{ANA}(⌜7⌝)⟩ ⊢ atom$} \RightLabel{Rule 2}\UnaryInfC{$⟨⌜7⌝, I_{ANA}(⌜7⌝)⟩ ⊢ n$} \RightLabel{Rule 3}\BinaryInfC{$⟨⌜37⌝, I_{ANA}(⌜3⌝) * 10^{length(⌜7⌝)} + I_{ANA}(⌜7⌝)⟩ ⊢ n$} \RightLabel{Rule 4}\UnaryInfC{$⟨⌜37⌝, I_{ANA}(⌜3⌝) * 10^{length(⌜7⌝)} + I_{ANA}(⌜7⌝)⟩ ⊢ np$} \AxiomC{$⟨⌜1⌝, I_{ANA}(⌜1⌝)⟩ ⊢ atom$} \RightLabel{Rule 2}\UnaryInfC{$⌜1⌝, I_{ANA}(⌜1⌝)⟩ ⊢ n$} \RightLabel{Rule 4}\UnaryInfC{$⌜1⌝, I_{ANA}(⌜1⌝)⟩ ⊢ np$} \RightLabel{Rule 5}\BinaryInfC{$⟨⌜(37 + 1)⌝, I_{ANA}(⌜3⌝) * 10^{length(⌜7⌝)} + I_{ANA}(⌜7⌝) + I_{ANA}(⌜1⌝)⟩ ⊢ np$} \AxiomC{$⟨⌜0⌝, I_{ANA}(⌜0⌝)⟩ ⊢ atom$} \RightLabel{Rule 2}\UnaryInfC{$⟨⌜0⌝, I_{ANA}(⌜0⌝)⟩ ⊢ n$} \RightLabel{Rule 4}\UnaryInfC{$⟨⌜0⌝, I_{ANA}(⌜0⌝)⟩ ⊢ np$} \RightLabel{Rule 5}\BinaryInfC{$⟨⌜((37 + 1) × 0)⌝, (I_{ANA}(⌜3⌝) * 10^{length(⌜7⌝)} + I_{ANA}(⌜7⌝) + I_{ANA}(⌜1⌝)) * I_{ANA}(⌜0⌝)⟩ ⊢ np$} \AxiomC{$⟨⌜5⌝, I_{ANA}(⌜5⌝)⟩ ⊢ atom$} \RightLabel{Rule 2}\UnaryInfC{$⟨⌜5⌝, I_{ANA}(⌜5⌝)⟩ ⊢ n$} \RightLabel{Rule 4}\UnaryInfC{$⟨⌜5⌝, I_{ANA}(⌜5⌝)⟩ ⊢ np$} \RightLabel{Rule 6}\BinaryInfC{$⟨⌜((37 + 1) × 0) = 5⌝, (I_{ANA}(⌜3⌝) * 10^{length(⌜7⌝)} + I_{ANA}(⌜7⌝) + I_{ANA}(⌜1⌝)) * I_{ANA}(⌜0⌝) = I_{ANA}(⌜5⌝)⟩ ⊢ s$} \end{prooftree}\]

Indeed, we’ve proved that $⌜((37 + 1) × 0 = 5⌝$ is an $s$. What we’ve also shown is that the interpretation of this $s$ is $\True$ just in case

\[(I_{ANA}(⌜3⌝) * 10^{length(⌜7⌝)} + I_{ANA}(⌜7⌝) + I_{ANA}(⌜1⌝)) * I_{ANA}(⌜0⌝)\]

is equal to $I_{ANA}(⌜5⌝)$.

In fact, since we know that \[I_{ANA}(⌜0⌝) = 0\] and that multiplying anything by $0$ gives you $0$, we can determine that

\[(I_{ANA}(⌜3⌝) * 10^{length(⌜7⌝)} + I_{ANA}(⌜7⌝) + I_{ANA}(⌜1⌝)) * I_{ANA}(⌜0⌝)\]

is equal to $0$ without even worrying about what

\[I_{ANA}(⌜3⌝) * 10^{length(⌜7⌝)} + I_{ANA}(⌜7⌝) + I_{ANA}(⌜1⌝)\]

evaluates to. Though, if we were to evaluate it, we’d of course find that it evaluates to $38$.

Since $I_{ANA}(⌜5⌝)$ is $5$, the interpretation of this sentence is $\True$ if and only if $0$ and $5$ are the same number. They’re not, so the interpretation is $\False$.

--- title: "Arabic numeral arithmetic: examples" bibliography: ../../semantics.bib format: html: css: ../styles.css html-math-method: mathjax mathjax-config: loader: {load: ['[tex]/bussproofs','[tex]/bbox','[tex]/colorbox']} tex: packages: {'[+]': ['bussproofs','bbox','colorbox']} --- ::: {.hidden} $$ \newcommand{\expr}[3]{\begin{array}{c} #1 \\ \bbox[lightblue,5px]{#2} \end{array} ⊢ #3} \newcommand{\ct}[1]{\bbox[font-size: 0.8em]{\mathsf{#1}}} \newcommand{\abbr}[1]{\bbox[transform: scale(0.95)]{\mathtt{#1}}} \def\True{\ct{T}} \def\False{\ct{F}} $$ ::: It's example time. ### Example 1 For the first example let's show that $⌜537⌝$ is an $np$ (and that it has a particular interpretation in $\mathcal{M}_{ANA}$). \begin{prooftree} \AxiomC{$⟨⌜5⌝, I_{ANA}(⌜5⌝)⟩ ⊢ atom$} \RightLabel{Rule 2}\UnaryInfC{$⟨⌜5⌝, I_{ANA}(⌜5⌝)⟩ ⊢ n$} \AxiomC{$⟨⌜3⌝, I_{ANA}(⌜3⌝)⟩ ⊢ atom$} \RightLabel{Rule 2}\UnaryInfC{$⟨⌜3⌝, I_{ANA}(⌜3⌝)⟩ ⊢ n$} \AxiomC{$⟨⌜7⌝, I_{ANA}(⌜7⌝)⟩ ⊢ atom$} \RightLabel{Rule 2}\UnaryInfC{$⟨⌜7⌝, I_{ANA}(⌜7⌝)⟩ ⊢ n$} \RightLabel{Rule 3}\BinaryInfC{$⟨⌜37⌝, I_{ANA}(⌜3⌝) * 10^{length(⌜7⌝)} + I_{ANA}(⌜7⌝)⟩ ⊢ n$} \RightLabel{Rule 3}\BinaryInfC{$⟨⌜537⌝, I_{ANA}(⌜5⌝) * 10^{length(⌜37⌝)} + I_{ANA}(⌜3⌝) * 10^{length(⌜7⌝)} + I_{ANA}(⌜7⌝)⟩ ⊢ n$} \RightLabel{Rule 4}\UnaryInfC{$⟨⌜537⌝, I_{ANA}(⌜5⌝) * 10^{length(⌜37⌝)} + I_{ANA}(⌜3⌝) * 10^{length(⌜7⌝)} + I_{ANA}(⌜7⌝)⟩ ⊢ np$} \end{prooftree} What's proved is the statement in the very last line of this derivation: that the string $⌜537⌝$ has the interpretation $$I_{ANA}(⌜5⌝) * 10^{length(⌜37⌝)} + I_{ANA}(⌜3⌝) * 10^{length(⌜7⌝)} + I_{ANA}(⌜7⌝)$$ and the category $np$. Since we know that - $I_{ANA}(⌜5⌝) = 5$, - $I_{ANA}(⌜3⌝) = 3$, and - $I_{ANA}(⌜7⌝) = 7$ as well as what $length(⌜37⌝)$ and $length(⌜7⌝)$ are---$2$ and $1$, respectively---the very last line of this proof/derivation can be rewritten as $$5 * 10^{2} + 3 * 10^{1} + 7$$ which, of course, is $537$. ### Example 2 For the second example, we'll show that $⌜((37 + 1) × 0) = 5⌝$ is an $s$, and also what its interpretation in $\mathcal{M}_{ANA}$ is. \begin{prooftree} \AxiomC{$⟨⌜3⌝, I_{ANA}(⌜3⌝)⟩ ⊢ atom$} \RightLabel{Rule 2}\UnaryInfC{$⟨⌜3⌝, I_{ANA}(⌜3⌝)⟩ ⊢ n$} \AxiomC{$⟨⌜7⌝, I_{ANA}(⌜7⌝)⟩ ⊢ atom$} \RightLabel{Rule 2}\UnaryInfC{$⟨⌜7⌝, I_{ANA}(⌜7⌝)⟩ ⊢ n$} \RightLabel{Rule 3}\BinaryInfC{$⟨⌜37⌝, I_{ANA}(⌜3⌝) * 10^{length(⌜7⌝)} + I_{ANA}(⌜7⌝)⟩ ⊢ n$} \RightLabel{Rule 4}\UnaryInfC{$⟨⌜37⌝, I_{ANA}(⌜3⌝) * 10^{length(⌜7⌝)} + I_{ANA}(⌜7⌝)⟩ ⊢ np$} \AxiomC{$⟨⌜1⌝, I_{ANA}(⌜1⌝)⟩ ⊢ atom$} \RightLabel{Rule 2}\UnaryInfC{$⌜1⌝, I_{ANA}(⌜1⌝)⟩ ⊢ n$} \RightLabel{Rule 4}\UnaryInfC{$⌜1⌝, I_{ANA}(⌜1⌝)⟩ ⊢ np$} \RightLabel{Rule 5}\BinaryInfC{$⟨⌜(37 + 1)⌝, I_{ANA}(⌜3⌝) * 10^{length(⌜7⌝)} + I_{ANA}(⌜7⌝) + I_{ANA}(⌜1⌝)⟩ ⊢ np$} \AxiomC{$⟨⌜0⌝, I_{ANA}(⌜0⌝)⟩ ⊢ atom$} \RightLabel{Rule 2}\UnaryInfC{$⟨⌜0⌝, I_{ANA}(⌜0⌝)⟩ ⊢ n$} \RightLabel{Rule 4}\UnaryInfC{$⟨⌜0⌝, I_{ANA}(⌜0⌝)⟩ ⊢ np$} \RightLabel{Rule 5}\BinaryInfC{$⟨⌜((37 + 1) × 0)⌝, (I_{ANA}(⌜3⌝) * 10^{length(⌜7⌝)} + I_{ANA}(⌜7⌝) + I_{ANA}(⌜1⌝)) * I_{ANA}(⌜0⌝)⟩ ⊢ np$} \AxiomC{$⟨⌜5⌝, I_{ANA}(⌜5⌝)⟩ ⊢ atom$} \RightLabel{Rule 2}\UnaryInfC{$⟨⌜5⌝, I_{ANA}(⌜5⌝)⟩ ⊢ n$} \RightLabel{Rule 4}\UnaryInfC{$⟨⌜5⌝, I_{ANA}(⌜5⌝)⟩ ⊢ np$} \RightLabel{Rule 6}\BinaryInfC{$⟨⌜((37 + 1) × 0) = 5⌝, (I_{ANA}(⌜3⌝) * 10^{length(⌜7⌝)} + I_{ANA}(⌜7⌝) + I_{ANA}(⌜1⌝)) * I_{ANA}(⌜0⌝) = I_{ANA}(⌜5⌝)⟩ ⊢ s$} \end{prooftree} Indeed, we've proved that $⌜((37 + 1) × 0 = 5⌝$ is an $s$. What we've also shown is that the interpretation of this $s$ is $\True$ just in case $$(I_{ANA}(⌜3⌝) * 10^{length(⌜7⌝)} + I_{ANA}(⌜7⌝) + I_{ANA}(⌜1⌝)) * I_{ANA}(⌜0⌝)$$ is equal to $I_{ANA}(⌜5⌝)$. In fact, since we know that $$I_{ANA}(⌜0⌝) = 0$$ and that multiplying anything by $0$ gives you $0$, we can determine that $$(I_{ANA}(⌜3⌝) * 10^{length(⌜7⌝)} + I_{ANA}(⌜7⌝) + I_{ANA}(⌜1⌝)) * I_{ANA}(⌜0⌝)$$ is equal to $0$ without even worrying about what $$I_{ANA}(⌜3⌝) * 10^{length(⌜7⌝)} + I_{ANA}(⌜7⌝) + I_{ANA}(⌜1⌝)$$ evaluates to. Though, if we were to evaluate it, we'd of course find that it evaluates to $38$. Since $I_{ANA}(⌜5⌝)$ is $5$, the interpretation of this sentence is $\True$ if and only if $0$ and $5$ are the same number. They're not, so the interpretation is $\False$.