Proving Invalidity in Predicate Logic by 29z84v

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```									 Proving Invalidity
in Predicate Logic

Kareem Khalifa
Department of Philosophy
Middlebury College
Overview
• Refresher on invalidity
• How this bears on predicate logic
• How to prove invalidity
– Play God!
– Rules for playing God
• Sample exercises
Refresher on invalidity
• An argument is invalid if there is some way
that all of its premises are true when its
conclusion is false.
Example of an invalid argument in
predicate logic
• Some philosophers have receding
hairlines.
• Therefore all philosophers have receding
hairlines.

+                  =
Invalidity in predicate logic
• An argument in predicate logic is invalid if
it there is some way all of its premises can
be true when its conclusion is false.
• However, we can’t quantify over
everything!
• So we get to…
– MAKE OUR OWN WORLDS!
– Though they’re kind of lame worlds…
Back to our example
• Imagine a world with two people, a and b,
both of whom are philosophers. However
a has a receding hairline and b does not.
• Then in this world, it would be true that
some philosophers have receding hairlines
(a does!) but false that all philosophers
have receding hairlines (b doesn’t!).
• So the argument is invalid.
Playing God
• To invalidate an argument, you get to create
very simple worlds, often consisting of no more
than two objects (let’s say a and b).
• You then use the following rules:
• Replace every statement of the form (x)(Фx),
with a disjunction Фa v Фb
• Replace every statement of the form (x)(Фx),
with a conjunction of the form Фa & Фb
• Now invalidate the argument as you would in
propositional logic with the disjunctions and
conjunctions.
Motivation for the
“Existential God Rule”

Let this be the whole wide world.   At least one thing is an F

So EITHER a is an F...

F
a                          F
b
a                         b            …OR b is an F.
Motivation for the
“Universal God Rule”

Let this be the whole wide world.     Everything is an F

So both a AND b are F’s.

a              F            b

a                  b
Formal treatment of our example
• (x)(Px & Rb)├ (x)(PxRx)  Rb)
(Pa & Ra) v (Pb &Rx)(Pa  Ra) & (Pb
– Let there be exactly two objects {a, b}
• Replace the (x) statements with v
statements.
• Replace your (x) statements with &
statements
• Now treat this just like a formal proof of
invalidity in propositional logic…
Continued
•    (Pa & Ra) v (Pb & Rb)├ (Pa  Ra) & (Pb  Rb)
1.   Let Pa, Pb, and Rb be true; Ra be false.
2.   Since Pa and Ra are false, Pa  Ra is false.
3.   Since Pa  Ra is false, the conclusion, (Pa  Ra) & (Pb  Rb)
is also false.
4.   Since Pb and Rb are true, Pb & Rb is true.
5.   Since Pb & Rb is true, the premise, (Pa & Ra) v (Pb & Rb), is also
true.
6.   By 3 and 5, all of the premises are true and the conclusion is
false.
7.   So the argument is invalid.

Pa Ra Pb      Rb    (Pa & Ra) v (Pb & Rb)       (Pa  Ra) & (Pb  Rb)
T    F   T    T     T                           F
How many objects should I put in
my world?
• Quick answer: As few as possible to prove
it invalid.
invalidate the argument, stop.
• If you can’t, create a two-object world. If
you can invalidate the argument, stop.
• If you can’t, create a three-object world…
Potential problem
• Suppose you don’t know whether the
argument is valid or invalid.
• Then this procedure could go on ad
infinitum for a valid argument!
Rules of thumb to avoid this
problem
• Try to sketch a proof of validity. If you “see the proof,”
then avoid playing God.
• In general, start by assuming the bare minimum required
to make either one premise true or a conclusion false.
Then fill in the rest as you need it.
• If all the premises and the conclusion are universally
quantified, then the argument can be invalidated using
only one object (467, A2).
• For each existentially quantified premise, posit a unique
object that makes exactly one premise true. (467, A5)
• When there are both universally and existentially
quantified statements in the argument, you generally
need 2+ objects. (467, A10)
(x)(Dx → ~Ex), (x)(Ex → Fx) ├
(x)(Fx → ~Dx)
• Da → ~Ea, Ea → Fa ├ Fa → ~Da
1. Let there be a model with domain {a}, such that Da and
Fa are true, and Ea is false.
2. Since Da is true, ~Da is false.
3. Since Fa is true and ~Da is false, the conclusion Fa →
~Da is false.
4. Since Ea is false, the premise Ea → Fa, is true.
5. Since Ea is false, ~Ea is true.
6. Since ~Ea is true, the premise Da → Ea is true.
7. Thus by 3, 4, and 6, it is possible for all of the premises
to be true and the conclusion to be false.
8. So the argument is invalid.
5. (x)(Mx & Nx), (x)(Mx & Ox) ├
(x)(Ox → Nx)
•    (Ma & Na) v (Mb & Nb), (Ma & Oa) v (Mb & Ob) ├ (Oa →
Na) & (Ob → Nb)
1.   Let there be a model {a, b} such that Ma, Na, Mb, and Ob
are true, and Nb is false.
2.   Since Ma and Na are true, Ma & Na is true.
3.   Since Ma & Na is true, the premise, (Ma & Na) v (Mb & Nb),
is true.
4.   Since Mb and Ob are true, Mb & Ob is true.
5.   Since Mb & Ob is true, the premise, (Ma & Oa) v (Mb &
Ob), is true.
6.   Since Ob is true and Nb is false, Ob → Nb is false.
7.   Since Ob → Nb is false, the conclusion (Oa → Na) & (Ob →
Nb) is false.
8.   By 3, 5, and 7, it is possible for all of the premises to be true
and the conclusion to be false.
9.   So the argument is invalid.
10. (x)(Bx & ~Cx), (x)(Dx → ~Cx)
├ (x)(Dx → Bx)
•     (Ba & ~Ca) v (Bb & ~Cb), (Da → ~Ca) & (Db → ~Cb) ├ (Da → Ba) & (Db
→ Bb)
1.    Let there be a model with domain {a, b} such that Ba and Db are true,
and Ca, Bb, and Cb are false.
2.    Since Ca is false, ~Ca is true.
3.    Since Ba and ~Ca are true, Ba & ~Ca is true.
4.    Since Ba & ~Ca is true, the premise (Ba & ~Ca) v (Bb & ~Cb) is true.
5.    Since Db is true and Bb is false, Db → Bb is false.
6.    Since Db → Bb is false, the conclusion (Da → Ba) & (Db → Bb)is false.
7.    By 2, ~Ca is true, so Da → ~Ca is true.
8.    Since Cb is false, ~Cb is true.
9.    Since ~Cb is true, Db → ~Cb is true.
10.   By 7 and 9, the premise (Da → ~Ca) & (Db → ~Cb) is true.
11.   Thus, by 4, 6, and 10, it is possible for all of the premises to be true and
the conclusion to be false.
12.   So the argument is invalid.

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