An English model and example

The model

Let’s now say a bit more about the model, $\mathcal{M}_{\textit{Eng.}}$.

$\mathcal{M}_{\textit{Eng.}} = ⟨D_{\textit{Eng.}}, \{\True, \False\}, I_{\textit{Eng.}}⟩$

The domain, $D_{\textit{Eng.}}$, we can take to be the set $\{\ct{z}, \ct{b}\}$:

\[D_{\textit{Eng.}} = \{\ct{z}, \ct{b}\}\]

Let us now also say a thing or two about the interpretation function, $I_{\textit{Eng.}}$. Consistent with our grammatical rules, we will:

interpret noun phrases as elements of the domain,
interpret verb phrases as characteristic functions of sets,
interpret auxiliaries as functions from characteristic functions to characteristic functions,
and interpret connectives as functions from pairs of characteristic functions to characteristic functions.

Here we go:

$I_{\textit{Eng.}}(\textit{Ziggy}) = \ct{z}$
$I_{\textit{Eng.}}(\textit{Bella}) = \ct{b}$
$I_{\textit{Eng.}}(\textit{runs}) = \{\ct{z}, \ct{b}\}_{CF}$
$I_{\textit{Eng.}}(\textit{run}) = \{\ct{z}, \ct{b}\}_{CF}$
$I_{\textit{Eng.}}(\textit{jumps}) = \{\ct{z}\}_{CF}$
$I_{\textit{Eng.}}(\textit{jump}) = \{\ct{z}\}_{CF}$
$I_{\textit{Eng.}}(\textit{doesn't}) = (λf.(λx.¬f(x)))$
$I_{\textit{Eng.}}(\textit{and}) = (λ⟨f, g⟩.(λx.f(x) = g(x) = \True))$

Recall the notation $(·)_{CF}$, which we use to form characteristic functions out of sets. Given any subset ($S$) of the domain $D$:

\[S_{CF} ≝ (λx.x ∈ S)\]

That is, a characteristic function of such a subset $D$ is a function whose domain is $D$ itself—$\{\ct{z}, \ct{b}\}$ above—and whose codomain is the set of truth values. Further, this function gives back $\True$ on any element of the domain which is a member of $S$ (otherwise, it gives back $\False$).

Also recall the “negation” function $¬ : \{\True, \False\} → \{\True, \False\}$. This function just flips $\True$ to $\False$ and $\False$ to $\True$:

\[¬\True = \False\] \[¬\False = \True\]

Alright. It is time to apply our lexicon definition and grammatical rules from earlier—together with this model definition—to an example expression of English.

Deriving an example

Here’s a derivation of the expression $\textit{Ziggy doesn't jump and runs}$. (Note that you may need to scroll right to see everything.)

\[\begin{prooftree} \AxiomC{$⟨\textit{Ziggy}, I_{\textit{Eng.}}(Ziggy)⟩ ⊢ np$} \AxiomC{$⟨\textit{doesn't}, I_{\textit{Eng.}}(doesn't)⟩ ⊢ aux$} \AxiomC{$⟨\textit{jump}, I_{\textit{Eng.}}(jump)⟩ ⊢ bvp$} \RightLabel{Rule 2}\BinaryInfC{$⟨\textit{doesn't jump}, I_{\textit{Eng.}}(doesn't)(I_{\textit{Eng.}}(jump))⟩ ⊢ vp$} \AxiomC{$⟨\textit{and}, I_{\textit{Eng.}}(and)⟩ ⊢ con$} \AxiomC{$⟨\textit{runs}, I_{\textit{Eng.}}(runs)⟩ ⊢ vp$} \RightLabel{Rule 3}\TrinaryInfC{$⟨\textit{doesn't jump and runs}, I_{\textit{Eng.}}(and)(I_{\textit{Eng.}}(doesn't)(I_{\textit{Eng.}}(jump)), I_{\textit{Eng.}}(runs))⟩ ⊢ vp$} \RightLabel{Rule 1}\BinaryInfC{$⟨\textit{Ziggy doesn't jump and runs}, I_{\textit{Eng.}}(and)(I_{\textit{Eng.}}(doesn't)(I_{\textit{Eng.}}(jump)), I_{\textit{Eng.}}(runs))(I_{\textit{Eng.}}(Ziggy))⟩ ⊢ s$} \end{prooftree}\]

Simplifying the result

Now, all we have to do is substitute the cases in the definition of $I_{\textit{Eng.}}$ in (2) for their occurrences in the last line of this derivation. If we do this we get:

\[(λ⟨f, g⟩.(λx.f(x) = g(x) = \True))((λf.(λx.¬f(x)))(\{\ct{z}\}_{CF}), \{\ct{z}, \ct{b}\}_{CF})(\ct{z})\]

This is a monster of a metalanguage expression! Fear not: it is quite straightforward to reduce it into something simple, as long as we are careful. One thing we need to be careful about is where all of the parentheses are!

Note that this expression can be divided up into three main parts. First, there is a function:

\[(λ⟨f, g⟩.(λx.f(x) = g(x) = \True))\]

That’s the first part. The second part is the function’s first argument, which is a pair:

\[⟨(λf.(λx.¬f(x)))(\{\ct{z}\}_{CF}), \{\ct{z}, \ct{b}\}_{CF}⟩\]

And the third part is the function’s second argument, which is just Ziggy:

\[\ct{z}\]

Importantly, the first argument of the function—the pair—is somewhat complex: it itself can be reduced into something simpler! Let’s talk about that.

Simplifying the first argument

The first argument of the main expression above

\[⟨(λf.(λx.¬f(x)))(\{\ct{z}\}_{CF}), \{\ct{z}, \ct{b}\}_{CF}⟩\]

is a pair, so it has two components. The first component of this pair is itself a function applied to an argument; specifically, it is the interpretation of doesn’t applied to the interpretation of jump:

\[(λf.(λx.¬f(x)))(\{\ct{z}\}_{CF})\]

Let us simplify this expression by first getting rid of the $λf$ at the beginning of the function and substituting the argument, $\{\ct{z}\}_{CF}$, for all occurrences of $f$ that appear inside of the function. Indeed, there is only one occurrence, and we get:

\[(λx.¬\{\ct{z}\}_{CF}(x))\]

Great! If we put this result back into the pair we started with, we end up with the following pair, which is equivalent to the one we needed to simplify:

\[⟨(λx.¬\{\ct{z}\}_{CF}(x)), \{\ct{z}, \ct{b}\}_{CF}⟩\]

Simplifying the main function applied to the first argument

If we apply the main function to its (now simplified) first argument, we obtain:

\[(λ⟨f, g⟩.(λx.f(x) = g(x) = \True))((λx.¬\{\ct{z}\}_{CF}(x)), \{\ct{z}, \ct{b}\}_{CF})\]

This complex metalanguage expression may be reduced as well. In particular, we should substitute

\[(λx.¬\{\ct{z}\}_{CF}(x))\]

in for the $f$ component of the function’s pair-shaped argument; and we should substitute

\[\{\ct{z}, \ct{b}\}_{CF}\]

for the $g$ component of this argument. As before, the substitution should occur inside the function itself, after we get rid of the $λ⟨f, g⟩$. If we do this complex substitution, we end up with

\[(λx.(λx.¬\{\ct{z}\}_{CF}(x))(x) = \{\ct{z}, \ct{b}\}_{CF}(x) = \True)\]

Having performed this substitution, we end up with another expression that can be simplified—specifically, the one to the left of the first equals sign:

\[(λx.¬\{\ct{z}\}_{CF}(x))(x)\]

This, too, is a function—$(λx.¬\{\ct{z}\}_{CF}(x))$—applied to an argument—$x$. In this case, we have to get rid of the $λx$ and—perhaps, unintuitively—substitute $x$ for the $x$ inside the function! No problem. If we do that, we get

\[¬\{\ct{z}\}_{CF}(x)\]

If we now try to recover the original function applied to its first argument—that is by replacing $(λx.¬\{\ct{z}\}_{CF}(x))(x)$ with its simplified version—we get

\[(λx.¬\{\ct{z}\}_{CF}(x) = \{\ct{z}, \ct{b}\}_{CF}(x) = \True)\]

It’s much easier to see now that this is just a characteristic function of some set: it is the characteristic function of the set of elements of the domain ($\{\ct{z}, \ct{b}\}$) such that it is true that that element is not in the set $\{\ct{z}\}$ and it is also true that element is in the set $\{\ct{z}, \ct{b}\}$. That’s just the characteristic function of the set containing $\ct{b}$—$\{\ct{b}\}_{CF}$!

Since this is the result of applying the main function to its first argument, we now have to apply the main function—already applied to its first argument—to its second argument.

Simplifying the main function (already applied to its first argument) applied to its second argument

Now that we have the result of applying the main function to its first argument, we can recover the original function applied to both its arguments by replacing the relevant part with the result we got. If we do this, we get

\[(λx.¬\{\ct{z}\}_{CF}(x) = \{\ct{z}, \ct{b}\}_{CF}(x) = \True)(\ct{z})\]

Thus, all we have remaining is one more substitution. In this case, we need to get rid of the $λx$ and substitute $\ct{z}$ inside the function, getting

\[¬\{\ct{z}\}_{CF}(\ct{z}) = \{\ct{z}, \ct{b}\}_{CF}(\ct{z}) = \True\]

This whole expression evaluates to $\True$ just in case $¬\{\ct{z}\}_{CF}(\ct{z})$ and $\{\ct{z}, \ct{b}\}_{CF}(\ct{z})$ are both true. Well… because $\{\ct{z}\}_{CF}(\ct{z})$ is $\True$, we have that $¬\{\ct{z}\}_{CF}(\ct{z})$ is $\False$. So, the meaning of Ziggy doesn’t jump and runs in $\mathcal{M}_{\textit{Eng.}}$ is $\False = \True = \True$. Otherwise, known as $\False$!

Final thoughts

It is worth going back up to the model definition and inspecting the meanings we provided for jump and runs. Hopefully, you can convince yourself that Ziggy doesn’t run and jumps really should be false—specifically, because Ziggy runs. I think it is then worth reflecting on the fact that we actually got this intuitive result in the end, simply by applying the rules of our grammar—carefully, but ultimately in a completely deterministic and mindless way—and then simplifying the resulting metalanguage expression.

--- title: "An English model and example" 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} $$ \newcommand{\expr}[3]{\begin{array}{c} #1 \\ \bbox[lightblue,5px]{#2} \end{array} ⊢ #3} \newcommand{\ct}[1]{\bbox[font-size: 0.8em]{\mathsf{#1}}} \newcommand{\abbr}[1]{\bbox[transform: scale(0.95)]{\mathtt{#1}}} \def\True{\ct{T}} \def\False{\ct{F}} $$ ::: #### The model Let's now say a bit more about the model, $\mathcal{M}_{\textit{Eng.}}$. (@ex-english-model-definition) $\mathcal{M}_{\textit{Eng.}} = ⟨D_{\textit{Eng.}}, \{\True, \False\}, I_{\textit{Eng.}}⟩$ The domain, $D_{\textit{Eng.}}$, we can take to be the set $\{\ct{z}, \ct{b}\}$: $$D_{\textit{Eng.}} = \{\ct{z}, \ct{b}\}$$ Let us now also say a thing or two about the interpretation function, $I_{\textit{Eng.}}$. Consistent with our grammatical rules, we will: - interpret noun phrases as elements of the domain, - interpret verb phrases as characteristic functions of sets, - interpret auxiliaries as functions from characteristic functions to characteristic functions, - and interpret connectives as functions from pairs of characteristic functions to characteristic functions. Here we go: (@ex-english-interp) $I_{\textit{Eng.}}(\textit{Ziggy}) = \ct{z}$ $I_{\textit{Eng.}}(\textit{Bella}) = \ct{b}$ $I_{\textit{Eng.}}(\textit{runs}) = \{\ct{z}, \ct{b}\}_{CF}$ $I_{\textit{Eng.}}(\textit{run}) = \{\ct{z}, \ct{b}\}_{CF}$ $I_{\textit{Eng.}}(\textit{jumps}) = \{\ct{z}\}_{CF}$ $I_{\textit{Eng.}}(\textit{jump}) = \{\ct{z}\}_{CF}$ $I_{\textit{Eng.}}(\textit{doesn't}) = (λf.(λx.¬f(x)))$ $I_{\textit{Eng.}}(\textit{and}) = (λ⟨f, g⟩.(λx.f(x) = g(x) = \True))$ Recall the notation $(·)_{CF}$, which we use to form characteristic functions out of sets. Given any subset ($S$) of the domain $D$: $$S_{CF} ≝ (λx.x ∈ S)$$ That is, a characteristic function of such a subset $D$ is a function whose domain is $D$ itself---$\{\ct{z}, \ct{b}\}$ above---and whose codomain is the set of truth values. Further, this function gives back $\True$ on any element of the domain which is a member of $S$ (otherwise, it gives back $\False$). Also recall the "negation" function $¬ : \{\True, \False\} → \{\True, \False\}$. This function just flips $\True$ to $\False$ and $\False$ to $\True$: $$¬\True = \False$$ $$¬\False = \True$$ Alright. It is time to apply our lexicon definition and grammatical rules from earlier---together with this model definition---to an example expression of English. #### Deriving an example Here's a derivation of the expression $\textit{Ziggy doesn't jump and runs}$. (Note that you may need to scroll right to see everything.) \begin{prooftree} \AxiomC{$⟨\textit{Ziggy}, I_{\textit{Eng.}}(Ziggy)⟩ ⊢ np$} \AxiomC{$⟨\textit{doesn't}, I_{\textit{Eng.}}(doesn't)⟩ ⊢ aux$} \AxiomC{$⟨\textit{jump}, I_{\textit{Eng.}}(jump)⟩ ⊢ bvp$} \RightLabel{Rule 2}\BinaryInfC{$⟨\textit{doesn't jump}, I_{\textit{Eng.}}(doesn't)(I_{\textit{Eng.}}(jump))⟩ ⊢ vp$} \AxiomC{$⟨\textit{and}, I_{\textit{Eng.}}(and)⟩ ⊢ con$} \AxiomC{$⟨\textit{runs}, I_{\textit{Eng.}}(runs)⟩ ⊢ vp$} \RightLabel{Rule 3}\TrinaryInfC{$⟨\textit{doesn't jump and runs}, I_{\textit{Eng.}}(and)(I_{\textit{Eng.}}(doesn't)(I_{\textit{Eng.}}(jump)), I_{\textit{Eng.}}(runs))⟩ ⊢ vp$} \RightLabel{Rule 1}\BinaryInfC{$⟨\textit{Ziggy doesn't jump and runs}, I_{\textit{Eng.}}(and)(I_{\textit{Eng.}}(doesn't)(I_{\textit{Eng.}}(jump)), I_{\textit{Eng.}}(runs))(I_{\textit{Eng.}}(Ziggy))⟩ ⊢ s$} \end{prooftree} #### Simplifying the result Now, all we have to do is substitute the cases in the definition of $I_{\textit{Eng.}}$ in (@ex-english-interp) for their occurrences in the last line of this derivation. If we do this we get: $$(λ⟨f, g⟩.(λx.f(x) = g(x) = \True))((λf.(λx.¬f(x)))(\{\ct{z}\}_{CF}), \{\ct{z}, \ct{b}\}_{CF})(\ct{z})$$ This is a monster of a metalanguage expression! Fear not: it is quite straightforward to reduce it into something simple, as long as we are careful. One thing we need to be careful about is where all of the parentheses are! Note that this expression can be divided up into three main parts. First, there is a *function*: $$(λ⟨f, g⟩.(λx.f(x) = g(x) = \True))$$ That's the first part. The second part is the function's *first argument*, which is a *pair*: $$⟨(λf.(λx.¬f(x)))(\{\ct{z}\}_{CF}), \{\ct{z}, \ct{b}\}_{CF}⟩$$ And the third part is the function's *second argument*, which is just Ziggy: $$\ct{z}$$ Importantly, the first argument of the function---the pair---is somewhat complex: it itself can be reduced into something simpler! Let's talk about that. #### Simplifying the first argument The first argument of the main expression above $$⟨(λf.(λx.¬f(x)))(\{\ct{z}\}_{CF}), \{\ct{z}, \ct{b}\}_{CF}⟩$$ is a pair, so it has two components. The first component of this pair is itself a function applied to an argument; specifically, it is the interpretation of *doesn't* applied to the interpretation of *jump*: $$(λf.(λx.¬f(x)))(\{\ct{z}\}_{CF})$$ Let us simplify this expression by first getting rid of the $λf$ at the beginning of the *function* and substituting the *argument*, $\{\ct{z}\}_{CF}$, for all occurrences of $f$ that appear inside of the function. Indeed, there is only one occurrence, and we get: $$(λx.¬\{\ct{z}\}_{CF}(x))$$ Great! If we put this result back into the pair we started with, we end up with the following pair, which is *equivalent* to the one we needed to simplify: $$⟨(λx.¬\{\ct{z}\}_{CF}(x)), \{\ct{z}, \ct{b}\}_{CF}⟩$$ #### Simplifying the main function applied to the first argument If we apply the main function to its (now simplified) first argument, we obtain: $$(λ⟨f, g⟩.(λx.f(x) = g(x) = \True))((λx.¬\{\ct{z}\}_{CF}(x)), \{\ct{z}, \ct{b}\}_{CF})$$ This complex metalanguage expression may be reduced as well. In particular, we should substitute $$(λx.¬\{\ct{z}\}_{CF}(x))$$ in for the $f$ component of the function's pair-shaped argument; and we should substitute $$\{\ct{z}, \ct{b}\}_{CF}$$ for the $g$ component of this argument. As before, the substitution should occur *inside* the function itself, after we get rid of the $λ⟨f, g⟩$. If we do this complex substitution, we end up with $$(λx.(λx.¬\{\ct{z}\}_{CF}(x))(x) = \{\ct{z}, \ct{b}\}_{CF}(x) = \True)$$ Having performed *this* substitution, we end up with *another* expression that can be simplified---specifically, the one to the left of the first equals sign: $$(λx.¬\{\ct{z}\}_{CF}(x))(x)$$ This, too, is a function---$(λx.¬\{\ct{z}\}_{CF}(x))$---applied to an argument---$x$. In this case, we have to get rid of the $λx$ and---perhaps, unintuitively---substitute $x$ for the $x$ inside the function! No problem. If we do that, we get $$¬\{\ct{z}\}_{CF}(x)$$ If we now try to recover the original function applied to its first argument---that is by replacing $(λx.¬\{\ct{z}\}_{CF}(x))(x)$ with its simplified version---we get $$(λx.¬\{\ct{z}\}_{CF}(x) = \{\ct{z}, \ct{b}\}_{CF}(x) = \True)$$ It's much easier to see now that this is just a characteristic function of some set: it is the characteristic function of the set of elements of the domain ($\{\ct{z}, \ct{b}\}$) such that it is *true* that that element is not in the set $\{\ct{z}\}$ and it is *also* true that element *is* in the set $\{\ct{z}, \ct{b}\}$. That's just the characteristic function of the set containing $\ct{b}$---$\{\ct{b}\}_{CF}$! Since this is the result of applying the main function to its first argument, we now have to apply the main function---already applied to its first argument---to its *second* argument. #### Simplifying the main function (already applied to its first argument) applied to its second argument Now that we have the result of applying the main function to its first argument, we can recover the original function applied to *both* its arguments by replacing the relevant part with the result we got. If we do this, we get $$(λx.¬\{\ct{z}\}_{CF}(x) = \{\ct{z}, \ct{b}\}_{CF}(x) = \True)(\ct{z})$$ Thus, all we have remaining is one more substitution. In this case, we need to get rid of the $λx$ and substitute $\ct{z}$ inside the function, getting $$¬\{\ct{z}\}_{CF}(\ct{z}) = \{\ct{z}, \ct{b}\}_{CF}(\ct{z}) = \True$$ This whole expression evaluates to $\True$ just in case $¬\{\ct{z}\}_{CF}(\ct{z})$ and $\{\ct{z}, \ct{b}\}_{CF}(\ct{z})$ are both true. Well... because $\{\ct{z}\}_{CF}(\ct{z})$ is $\True$, we have that $¬\{\ct{z}\}_{CF}(\ct{z})$ is $\False$. So, the meaning of *Ziggy doesn't jump and runs* in $\mathcal{M}_{\textit{Eng.}}$ is $\False = \True = \True$. Otherwise, known as $\False$! #### Final thoughts It is worth going back up to the model definition and inspecting the meanings we provided for *jump* and *runs*. Hopefully, you can convince yourself that *Ziggy doesn't run and jumps* really [should]{.underline} be false---specifically, because Ziggy runs. I think it is then worth reflecting on the fact that we actually got this intuitive result in the end, simply by applying the rules of our grammar---carefully, but ultimately in a completely deterministic and mindless way---and then simplifying the resulting metalanguage expression.