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Philosophy of Language

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Title: Philosophy of Language


1
lecture 9
  • Philosophy of Language

2
  • Intuition
  • (1) George wishes to know whether Scott is the
    author of Waverley
  • is true, and
  • (2) George wishes to know whether Scott is Scott
  • is false.
  • One reason why Intuition seems problematic it
    seems to violate Leibnizs Law (co-referential
    terms are substitutable salva veritate)
  • Russells reply (2) does not follow from (1) by
    LL (even though Scott is the author of Waverley)

3
  • Why is the argument
  • P1 G wtkw Scott is the author of W
  • P2 the author of W is Scott
  • P3 G wtkw Scott is Scott
  • not something to which LL applies?
  • Russell because it is not of the form Fa, ab,
    thus Fb
  • That is because P1 is not of the form Fa and P2
    is not of the form ab
  • That is because the author of W is not of the
    form a (it is not a singular term)

4
  • This ends Russells solution to the puzzle of
    George as a counter-example to LL
  • But there could be more to the puzzle all right,
    C does not follow from P1 and P2 by virtue of LL,
    but we ought to make sure it does not follow by
    virtue of other logical laws either
  • I.e. we have true premises and a false
    conclusion, so any analysis of these sentences
    should be such that it invalidates the argument
    (i.e., no logical rule can allow the step from P1
    and P2 to C)
  • So, Russell needs to make sure that his analysis
    of the author of W does not yield a logical
    form for P1-P2-C that is valid. Otherwise, LL or
    not, a false conclusion would follow from true
    premises.

5
  • Here is were the idea of scope ambiguities comes
    in. Russell says
  • P1 is scope-ambiguous it is true on one reading
    (quantifier with narrow scope) but false in
    another (quantifier with wide scope)
  • I.e., P1 has two logical forms, two readings
  • There is a way of deriving the false conclusion
    from P2 and one of the readings of P1, but that
    is not a problem, since that reading is false
  • If P1 is given the true reading, then even if
    coupled with the true P2, it does not yield the
    false conclusion

6
  • A bit more detail
  • 1. ?x author(x) ?y(author(y) ? yx)
    GWTKW(xScott) (Prem)
  • 2. ?x author(x) ?y(author(y) ? yx)
    xScott (Prem)

7
  • A bit more detail
  • 1. ?x author(x) ?y(author(y) ? yx)
    GWTKW(xScott) (Prem)
  • 2. ?x author(x) ?y(author(y) ? yx)
    xScott (Prem)
  • 3. author(a) ?y(author(y) ? ya)
    GWTKW(aScott) 1.EG

8
  • A bit more detail
  • 1. ?x author(x) ?y(author(y) ? yx)
    GWTKW(xScott) (Prem)
  • 2. ?x author(x) ?y(author(y) ? yx)
    xScott (Prem)
  • 3. author(a) ?y(author(y) ? ya)
    GWTKW(aScott) 1.EG
  • 4. author(b) ?y(author(y) ? yb) b
    Scott 2. EG

9
  • A bit more detail
  • 1. ?x author(x) ?y(author(y) ? yx)
    GWTKW(xScott) (Prem)
  • 2. ?x author(x) ?y(author(y) ? yx)
    xScott (Prem)
  • 3. author(a) ?y(author(y) ? ya)
    GWTKW(aScott) 1.EG
  • 4. author(b) ?y(author(y) ? yb) b
    Scott 2. EG
  • 5. ?y(author(y) ? ya) 3.

10
  • A bit more detail
  • 1. ?x author(x) ?y(author(y) ? yx)
    GWTKW(xScott) (Prem)
  • 2. ?x author(x) ?y(author(y) ? yx)
    xScott (Prem)
  • 3. author(a) ?y(author(y) ? ya)
    GWTKW(aScott) 1.EG
  • 4. author(b) ?y(author(y) ? yb) b
    Scott 2. EG
  • 5. ?y(author(y) ? ya) 3.
  • 6. author(b) ? ba 5, UI

11
  • A bit more detail
  • 1. ?x author(x) ?y(author(y) ? yx)
    GWTKW(xScott) (Prem)
  • 2. ?x author(x) ?y(author(y) ? yx)
    xScott (Prem)
  • 3. author(a) ?y(author(y) ? ya)
    GWTKW(aScott) 1.EG
  • 4. author(b) ?y(author(y) ? yb) b
    Scott 2. EG
  • 5. ?y(author(y) ? ya) 3.
  • 6. author(b) ? ba 5, UI
  • 7. ba 4.,6.

12
  • A bit more detail
  • 1. ?x author(x) ?y(author(y) ? yx)
    GWTKW(xScott) (Prem)
  • 2. ?x author(x) ?y(author(y) ? yx)
    xScott (Prem)
  • 3. author(a) ?y(author(y) ? ya)
    GWTKW(aScott) 1.EG
  • 4. author(b) ?y(author(y) ? yb) b
    Scott 2. EG
  • 5. ?y(author(y) ? ya) 3.
  • 6. author(b) ? ba 5, UI
  • 7. ba 4.,6.
  • 8. a Scott 4., 7. LL

13
  • A bit more detail
  • 1. ?x author(x) ?y(author(y) ? yx)
    GWTKW(xScott) (Prem)
  • 2. ?x author(x) ?y(author(y) ? yx)
    xScott (Prem)
  • 3. author(a) ?y(author(y) ? ya)
    GWTKW(aScott) 1.EG
  • 4. author(b) ?y(author(y) ? yb) b
    Scott 2. EG
  • 5. ?y(author(y) ? ya) 3.
  • 6. author(b) ? ba 5, UI
  • 7. ba 4.,6.
  • 8. a Scott 4., 7. LL
  • 9. GWTKN(ScottScott) 3.8. LL

14
  • A bit more detail
  • 1. ?x author(x) ?y(author(y) ? yx)
    GWTKW(xScott) (Prem)
  • 2. ?x author(x) ?y(author(y) ? yx)
    xScott (Prem)
  • 3. author(a) ?y(author(y) ? ya)
    GWTKW(aScott) 1.EG
  • 4. author(b) ?y(author(y) ? yb) b
    Scott 2. EG
  • 5. ?y(author(y) ? ya) 3.
  • 6. author(b) ? ba 5, UI
  • 7. ba 4.,6.
  • 8. a Scott 4., 7. LL
  • 9. GWTKN(ScottScott) 3.8. LL

15
  • This is a valid argument, but this much is not
    problematic since it starts with a false premise.
    NB assuming that GWTKN has not some special
    properties in fact it does, it is an
    hyperintensional operator, but Russells solution
    does not appeal to this
  • When we look at things from the surface (i.e.,
    focusing not on logical forms but on English
    sentences), this argument summarized what Russell
    calls verbal substitution.
  • NB this is not a logical law (surface English
    has no logic)
  • But it is a way of expressing a series of valid
    steps.
  • Russells point verbal substitution holds (in
    the case of George) only if the description is
    given wide scope. This move (i.e., this series of
    logical steps) is however applicable only when
    the description has wide scope.

16
  • So a real logical law, LL, is never violated
    by the story of G (simply because we do not have
    two singular terms)
  • There is a group of steps that apply to the
    argument under one reading, verbal
    substitution, and this is also not violated
  • We do not find a counterexample to verbal
    substitutivity, because if it applies, the
    premise is false, and if the premise is true, it
    does not apply

17
  • One step left for Russell show that, if P1 is
    read with narrow scope quantifier (i.e., as
    true), there is no other way (LL, verbal
    substitution, or whatever) to go from premises to
    conclusion.
  • I.e., that
  • GWTKW ?x(author(x) ?y(author(y)?yx)
    xScott
  • ?xauthor(x) ?y(author(y)?yx) xScott
  • GWTKW ScottScott
  • is invalid.
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