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I'm currently in a mathematical linguistics course, and I'm having trouble understanding the meaning of

'g[d/v]: the variable assignment g′ that is exactly like g except (maybe) for g(v), which equals the individual d'

in the semantics of predicate logic. If given a variable assignment in an example model, does this mean that (v) refers to all variables (d) that is in the universe, thus that all elements of the universe replaces the variable (v)?

g1 =

x1 → John

x2 → Mary

x3 → Pete

xn → Pete (where n≥4)

g1[John/x3] =

x1 → John

x2 → Mary

x3 → John

xn → Pete (where n≥4)

g1[[John/x3]Pete/x1] =

x1 → Pete

x2 → Mary

x3 → John

xn → Pete (where n≥4)

Also, I have an exercise based on variable assignment equivalence, but I do not know how to approach answering these questions since I do not entirely understand the meaning of variable assignment, and modified variable assignment.

QUESTION: Complete the equivalences assuming: g(x) = Mary, and g(y) = Susan.

1. g[Paul/x)(x) =

2. g[Paul/x)(y) =

3. g[[Paul/x]Susan/x)(x) =

4. g[[Paul/x]Susan/x)(y) =

5. g[(Paul/x)Susan/y)(x) =

6. g[[Paul/x]Susan/y)(y) =

If anyone could explain this concept to me, I would be very grateful!

EDIT: sorry, i'm quite new to this site! the questions were cut off by the closed bracket. I've tried attempting the questions below.

1.g[Paul/x)(x) = x: Paul

2.g[Paul/x)(y) = y: Susan

3.g[[Paul/x]Susan/x)(x) = x: Susan?

4.g[[Paul/x]Susan/x)(y) = y: Susan?

5.g[[Paul/x]Susan/y)(x) = x: Paul?

6.g[[Paul/x]Susan/y)(y) = y: Susan?

I'm a bit unsure about some of these, if x is originally mapped to Mary, then to Paul & Susan in (3&4)

LizJu
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2 Answers2

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If given a variable assignment in an example model, does this mean that (v) refers to all variables (d) that is in the universe, thus that all elements of the universe replaces the variable (v)?

No. A variable assignment maps every variable to a specific individual. You can see that with your first example:

g1 =

x1 → John

x2 → Mary

x3 → Pete

xn → Pete (where n≥4)

However, we can change those assignments when we do something like:

g1[John/x3]

This means that everything gets assigned the same individual as above, except that we now map $x3$ to John, so we get:

g1[John/x3] =

x1 → John

x2 → Mary

x3 → John

xn → Pete (where n≥4)

So, for last exercise at the end, your initial g is:

g =

x → Mary

y → Susan

So that means that g[Paul/x] is:

g[Paul/x] =

x → Paul

y → Susan

Can you do the others?

Bram28
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  • Sorry! I didn't realize that my questions were cut off! So would these be the answers?

    1.gPaul/x = x: Paul 2.gPaul/x = y: Susan 3.g[Paul/x]Susan/x = x: Susan? 4.g[Paul/x]Susan/x = y: Susan? 5.g[Paul/x]Susan/y = x: Paul? 6.g[Paul/x]Susan/y = y: Susan?

    I'm a bit unsure about some of these, if x is originally mapped to Mary, then to Paul & Susan in (3&4)

    – LizJu Jul 15 '17 at 20:08
  • @LizJu I am a little unsure of those myself ... it's clear what g[d/x] is, but I am not sure how to interpret something like g[[d/x]c/x] ... does your book have an example or maybe a precise definition for this notation? – Bram28 Jul 15 '17 at 22:38
  • There is a definition & examples listed from my book, pg. 126-128, but I am unable to understand it.. the ebook version: https://www.phil-fak.uni-duesseldorf.de/fileadmin/Redaktion/Institute/Allgemeine_Sprachwissenschaft/Dokumente/Bilder/01_Chierchia___McConnell-Ginet__2000_.pdf – LizJu Jul 15 '17 at 23:57
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    @LizJu OK, thanks. I look edt aht pages and udnerstand it now: the inside assignment is the first modification, and the outside one modifies the inside one. So, for example, when you take g[[Paul/x]Susan/x], you take the original g, then modify so as to map x to Paul, and then modifi it again to have x point to Susan. Hence, g[Paul/x]Susan/x = Susan. In fact, you all of them correct! – Bram28 Jul 16 '17 at 03:00
  • Ohhh, that makes so much more sense!! I noticed you answered my other post as well, thank you so much taking time to clarify these concepts for me!! :) – LizJu Jul 16 '17 at 06:07
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A variable assignment assigns a unique value to every variable. Think of it as a list

$[v_1 \mapsto d_1, v_2 \mapsto d_2 \ldots ]$

that describes that variable $v_1$ has value $d_1$, variable $v_2$ has value $d_2$ etc.

If $g$ is a variable assignment, then the updated variable assignment $g[d/v]$ is the same assignment except when it comes to the variable $v$. The list representation of $g[d/v]$ is the same as that of $g$ except in one place, namely that for $v$.

If $g$ is the list

$[v_1 \mapsto d_1, v_2 \mapsto d_2, v \mapsto d' \ldots ]$

then $g[d/v]$ is the list

$[v_1 \mapsto d_1, v_2 \mapsto d_2, v \mapsto d \ldots ]$.

Hans Hüttel
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