We have already seen how we can eliminate the arrow with the rules MP
and MT, but we have not seen a corresponding rule to introduce
it. We can often enough find a way of establishing a disjunction
and then using the implication rule to convert it to a conditional.
P & Q |- P -> Q
1. P & Q /
show Q 2.
Q 1,
Simp 3. ~P v Q
2, Add 4. P ->
Q 3, Impl
note that we can do
something like this whenever we know that the consequent of the
intended consitional is true or that the antecedent is false
But we also have a new and often quite efficient derivation
technique. Conditional or
hypothetical derivations give us a way to
establish the truth of a conditional (in other words, allowing us to
introduce the arrow). The key to understanding this is that if we
can show that a particular consequent would have to follow once we know
a given antecedent then we can say the conditional relationship exists,
regardless of whether the antecedent really is true. Again, as we
did with indirect derivations, we indent for a subordinate proof, and
we indicate the intention of the hypothesis. Here we say
"HCP" to indicate this is a hypothesis in a conditional proof (CP).
Here are some very simple examples:
P -> Q, Q -> R |- P -> R
1. P -> Q 2. Q -> R
/show P -> R 3.
P
HCP we hypothesize that the antecedent P is true 4.
Q 1,3, MP 5.
R 2,4, MP 6. P ->
R 3-5, CP we show that we have discharged the
hypothesis
|- (~P v Q) -> (P
-> Q)
1. ~P v
Q HCP
2. P
HCP since the
consequent is itself a conditional, we can bring in another subordinate
derivation
3. Q 1,2, DS 4. P
-> Q 2-3, CP 5. (~P v Q) -> (P ->
Q) 1-4, CP
In working with quantified
expressions we often find an exception to the usual rule about not
introducing a universal quantifier with a name that did not first come
from a statement for which the main connective was a universal
quantifier.
(Ex)Fx -> (y)Gy |- (x)Fx -> Gx)
1.
(Ex)Fx -> (y)Gy / show (x)(Fx -> Gx)
2. Fa
HCP
3. (Ex)Fx 2, EG
4. (y)Gy 1,3, MP
5.
Ga
4, UI 6.
Fa -> Ga 2-5, CP 7.
(x)(Fx -> Gx) 6, UG Now if this last step seems as though we
are saying something about everyone when we only started with a
particular name (Al's , let's say), the reason it works is that the
hypothesis itself could have assumed any name at all and produced the
same result. We need not even have to assume that there is anyone
meets the condition of being F so as to let us know that the same
individual would also have be G.
EXERCISES
(ON YOUR OWN)
Symbolize each of the
following arguments (both of which are intended
to be valid) and use conditional derivations.
1. If Alice studies then
she will get most of the answers
correct, and if she gets most of the answers correct then she will pass
the test. Therefore, if Alice does not pass the test then she did
not study. [use S,A,P] 2. Knowing that Alice
studies only if she is not working and she
is not working only if she does not have a job proves that unless she
is not studying she does not have a job. [use S,W,J]
REVIEW
EXERCISES FOR BONUS CREDIT (2 POINTS) Copy and
paste the arguments and your work (symbolization in predicate logic and
conditional proofs) into an email to me at dmcf34@yahoo.com no later
than December 12
1. If
Alice is ambitious then she is a good student. All good students
are happy. Therefore, if Alice is a good student she is happy. 2. No
clowns are sad. Therefore, if Bozo is a clown he is not sad. 3.
Unless they are lazy all students will work hard. Anyone who
works hard will pass. Therefore, if Alice is a student and she is
not lazy we can be sure she will pass. 4. If
there is someone winning then everyone will be happy. Therefore,
anyone who wins will be happy.
