# IDENTITY

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```					 “ It all depends
on what you mean
by “is” ”

President Clinton

1
IDENTITY

Dr Jekyll is Mr Hyde.

The Prime Minister is
Tony Blair.

Everest is
Chomdongolinga.

n=m.
2
Tricky Identities

Icabod is the smartest
student in the class.
i
1= - e

Sure-Fire Identities

Icabod is Icabod.

n=n
3
Tableau for
Quantifiers.

Rule :

Whenever [d=d]
occurs on a branch,
close that branch.

4
Dobaci is even smarter than
the tallest person in the room.
Dobaci is Icabod
 Icabod is even smarter than
the tallest person in the room

i: Icabod
d: Dobaci
Sx: x is even smarter than
the tallest person in the room

Sd, d=i  Si
5
n: Hyde
m : Jekyll

Tx : x is tall

Tn, n=m         Tm

Seems OK

6
Leibniz’ Rule

Leibniz 1646 - 1716

Indiscernibility of
Identicals.

But Shakespeare had
it first.

Shakespeare
1564 - 1616
7
Tableau

.
.
.
Tn
n=m
|
Tm

8
Shakespeare in Trouble !

Tn, n=m      Tm

Does not always work !

9
Lxy : x was in love with y
n : The Master of Balliol
m : Andrew Graham
f : Florence Nightingale

Lnf,    n=m      Lmf

As so often in affairs of the
heart, something has gone
wrong.

10
1) Lnf

2) n=m

“n” designates different
things in 1) and 2)

1) Jowett

2) Graham

11
Build Leibniz/
Shakespeare into our
logic.

Exclude any cases in
which it fails.

My policy : don’t go
where Leibniz fears
to follow.

12
NS believes that Marseille
is the capital of France.

Paris is the capital of
France.

 NS believes that
Marseille is Paris

Propositional attitude
(intentional contexts)

13
It could have been the
Margaret Thatcher is the
Prime Minister.

The Prime Minister is
Tony Blair

It could have been that
Margaret Thatcher is
Tony Blair.

invalid.

Modal contexts.
14
Problems for Leibniz

1. Temporal shifts

2. Propositional
attitudes

3. Modalities

We simply set these asides
for a later occasion.

15
Hodge’s Policy :
second assumption
we assume every
designator is purely
referential

Test : re-write with the
designator at the front

The Master of Balliol is
such that he was in love
with Florence.
16
If it expresses the same as
the original, the
designator is purely
referential.

In the front, it picks out
the primary referent (the
shopping list referent).

So this case is not purely
referential.

17
The Prime Minister is
such that he/she
could have been
Margaret Thatcher.

Difference in what it
expresses. So not
purely referential.

obsurities in this
procedure (expresses
the same thing?)
18
The Honest Approach

avoid arguments in which
Leibniz law fails.

19
Hodges First Assumption

When dealing with sentences
in which designator is purely
referential, we assume that it
has a primary referent.

NS: don’t attempt to deal
with designators which do
not have a primary referent.

In particular, avoid vacuous
names.

20
Tableau for Quantifiers

xFx
|
Fn

NB : “n” must be an
OLD name.

“n” must already be
in play in the tableau.

WEIRD
21
xFx
|
Fn
NB : “n” must be a new
name.

“n” must not have
already occured in the
tableau.

This is fair enough as
we will see.
22
xFx
|
xFx

If it is false that
everyone is funny,
then you can find
someone who is not
funny.
23
xFx
|
xFx

If it is false that you
can find someone
who is funny, then
take anyone you like
he or she will not be
funny.
24
Sx : x is a student
Rx : x is rich
n : Icabod

Domain : residents of
Oxford

Sn, Rn x[Sx  Rx]

CES
{Sn,Rn, x[Sx  Rx]}
25
Sn, Rn      x(Sx  Rx)

Sn
Rn
x(Sx  Rx)
|
x(Sx  Rx)
|
(Sn  Rn)

Sn       Rn
tableau closes, argument
valid
26
Cx : x is at Christ
Church
Rx : x is rich
n : Icabod
Domain : students in
Oxford

Cn, x(Cx  Rx)   Rn

CES

{Cn, x(Cx  Rx), Rn}
27
Cn, x(Cx  Rx) Rn

Cn
x(Cx  Rx)
Rn
|
(Cn  Rn)

Cn     Rn

closes, argument valid
28
Rx : x is rich
Sx : x is sad
Domain : those in this
room

xRx, xSx x(Rx  Sx)

CES

{xRx, xSx, x(RxSx)}

29
?
xRx, xSx x(Rx  Sx)
xRx
xSx
x(Rx  Sx)
|
x(Rx  Sx)
Ra
!!!   Sa
|
(Ra  Sa)
Ra         Sa
!!! closes but not valid so use
new names
30
Theorem

something you can prove
without using any
assumptions.

PP

CES    { [P  P] }

[P  P]
P
P
P
31
Theorems represent
truths of logic. They
are true come what
may.

x [ Fx  Fx]

32
Something green
exists.

xGx

Something exists.

x[x = x]

Cannot just write : x

not even a sentence
33
Consider whether it
should be a theorem of
logic that x[x = x]

CES      { x[x = x] }

x[x = x]
|
x[x = x]

end of the road for Hodges

NS forges on :
[b = b]
did not use an old name.
34
Rx: x is rich
Hx : x is happy
domain: this audience

xRx, x[RxHx] xHx

CES
{xRx, x[Rx  Hx], xHx}

35
xRx
x[Rx  Hx]
xHx
|
xHx
|
Ra
|
[Ra  Ha]

Ra       Ha
|
Ha

closes, therefore valid.
36

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