## 2014-07-15

### Quantification

$\newcommand{\R}{\mathbb R}$This follows on from my previous language post, and is not about mathematical quantification specifically, though it bears resemblance to what Jakub Marian does. As before, I'm trying to use Lojban words but with new grammar. This, however, is not what I expect this unnamed language to be.

You may remember that I set {su'o} (“at least 1”) as the default quantifier. I also choose to have suffix quantifiers, so 〈gerku fafa plipe〉 = 〈gerku fasu'o fasu'o plipe〉. The way I imagine this is that there is a list of things matching 〈gerku fa〉 (dogs) on one side, and a list of things matching 〈plipe fa〉 (leapers) on the other. Any identical things from opposite sides are highlighted. There are at least 1 things highlighted on the 〈gerku fa〉 side, and at least 1 things highlighted on the 〈plipe fa〉 side.

The second quantifier may seem redundant. Can't we just draw lines between matching items from opposite sides, then count the lines? The answer is “no” because some quantifiers take into account properties of a single referent set. This is most obviously seen with {ro} - “every”. 〈gerku faro fa plipe〉 says that every dog leaps, and 〈gerku fa faro plipe〉 says that every leaper is a dog. By the way, both are false in this universe, but the former leads to an interesting topic that I'll discuss later.

There also exists 〈gerku faro faro plipe〉, which states that the referent sets of 〈gerku fa〉 and 〈plipe fa〉 are identical. We can see this as an alternative to {jo} (“if and only if”), but based on set membership rather than basic logic. This reasoning carries over to the previous two examples, where 〈gerku faro fa plipe〉 means “x is a dog implies that x leaps”. I find this much clearer than Lojban's {da gerku nagi'a plipe}, and I will strive to make the quantifier method available in situations that currently require basic logic in Lojban. Here, of course, {ro gerku cu plipe} would suffice.

A point I haven't covered is about order of quantification. In maths, $\forall x\in\R.\exists y\in\R.x+1=y$ and $\exists y\in\R.\forall x\in\R.x+1=y$ are very different claims, with the first being true and the second being false. So, how do we translate these into the new language?

First, we need some vocabulary. So, let 〈+〉 stand for {sumji} (“fa is the sum of fe and fi”), and 〈1〉 and 〈R〉 stand for the predicates “fa = 1” and “fa ∈ ℝ”, respectively. 〈pa〉 acts as a quantifier meaning “there is exactly one”. We also need some more grammar, so let 〈)〉 stand for “hide the previous selbri from now on”. This helps in translating infix notation. So, the first claim becomes 〈R faro fero + firo fa 1 ) faro fa R〉 and the second becomes 〈R faro fero + firo fa 1 ) faro fapa R〉 (the 〈fa〉s after the 〈)〉s are referring to a place of the 〈+〉s).

These bear little resemblance to the original claims, so take some thinking to understand (and to write). The logic is completely different, since the new language is based on comparing referent sets rather than finding solutions. The first claim essentially says $\R+1\subseteq\R$, whereas the second is $\{y\}=\R+1\land y\in\R$. The second is wrong because we're trying to squeeze all the results of $\R+1$ into a single element.

But wait, “there exists” is not the same as “there exists unique”. Have we lost that distinction? Let's test with $\forall x\in\R.\exists y\in\R.{}_\cdot^2x=y$ and $\forall x\in\R.\exists y\in\mathbb C.x={}_\cdot^2y$, both of which should be true. We can translate these into set equations as we did for the previous examples, giving ${}_\cdot^2\R\subseteq\R$ and $\R\subseteq{}_\cdot^2\mathbb C$. Let 〈kurtenfa〉 be “fa is the square of fe”, and 〈C〉 be “fa ∈ ℂ”. All of that gives us 〈R faro fero kurtenfa faro fa R〉 and 〈R faro faro kurtenfa fero fa C〉. They have no references to explicit numerical quantities, so we have avoided claiming that “there exists” = “there exists unique”.

So, where did the 〈pa〉 come from? Note first of all that we were trying to make a false claim, which may have caused the strange form. But thinking about it, what we were claiming with $\exists y\in\R.\forall x\in\R.x+1=y$ was that ${\large\iota}\in\R\implies{\large\iota}+1=\text{const}\ y$, where $y$ itself is constant. $\text{const}\ y$ has only one value it can return, hence the 〈pa〉.