Topic DS: How to prove things with Sentential logic

3 More rules

3.1 Rules for ''

As for '&' and '→', there is both an elimination rule
and an introduction rule for '

I (Disjunction Introduction)
If you have derived φ,
you can write down (φ
ψ) or (ψφ),
depending on everything φ depends on.

E (Disjunction Elimination or Disjunctive Syllogism)
If you have derived (φ
ψ) and ~ψ,
you can write down φ,
depending on everything (φ
ψ) and ~ψ depend on.
If you have derived (φ
ψ) and ~φ,
you can write down ψ,
depending on everything (φ
ψ) and ~φ depend on.

The introduction rule says that you can take any
formula you have written down so far and make it
longer, by changing it into a disjunction. And you
write down the same dependencies. So if the original
formula is
φ, then the new formula would beψ) or
φ). Note that ψ can be any formula at all, from simple,
like "A" or "B" to more complicated, like "(A & B)" or
"((A→B)&(B→A))". For example


The elimination rule for '' lets you break apart a
disjunction and write down only one half. Like &I,
E starts from two formulas and permits you to write
down a third.
E permits you to write down φ if you
ψ) and in your derivation. E also permits
you to write down
ψ, if you haveψ) and in your
derivation. The dependencies of the new formula are
the same as the dependencies of two the formulas you
had already. A more traditional name for this sort of rule
Disjunctive Syllogism; we will just call it Elimination,
E for short.

For example:


At this point you can show the interesting
fact that anything is derivable from an explicit
contradiction: for any
φ and ψ, (φ&~φ) proves ψ.
For example, (B & ~B)
proves A.
Can you see how to show that?

Exercise 3.1a

Show (B & ~B)
proves A.

There is one more rule involving '
', Proof
by Cases
, or PC:

PC (Proof by Cases)
If you have derived (φ
ψ) and (φα) and (ψβ),
then you can write down (α
depending on everything (φ
ψ) and
α) and (ψβ) depend on.

It may look complicated, but it is actually almost
as simple as &I or
E. You need three formulas
in order to apply the rule. Sometimes it may be
difficult to see how to get these three formulas.
But if you do have them written in your derivation,
then you can write down a certain disjunction.
The new formula depends on everything the three
formulas you started with depend on.

Exercise 3.1b

Can PC be applied, starting from only 2
different formulas?

Exercise 3.1c
Show ((P & P)
(Q & Q)) proves (P Q)

Exercise 3.1d
Show (P
P), (P → Q) proves (P Q)

3.2 Rules for '~'

There are just two more rules left,
the rules for the introduction and elimination
of the negation symbol '~'.

Rules ~I and ~E are similar to →I in certain ways.
Each of these three rules permits you to take
away dependencies. In addition, with each of these
three rules, you first assume something, and then
you derive something, and then you apply the rule
to write down a new formula.

Here is Rule ~I:

~I (Negation Introduction)
If you have assumed ψ, and you have derived (φ&~φ),
then you can write down ~ψ,
depending on everything (φ&~φ) depends on except ψ.

Rule ~I provides one way to derive a negation
like "~A" or "~(A&B)". Suppose, for example, you want
to derive "~(A&B)". In using ~I you would first assume
"(A&B)" and then derive an explicit contradiction.
(An explicit contradiction is a formula of the form
(φ&~φ) such as "(A&~A)" or "((BC)&~(BC))".)
Then you can write down "~(A&B)".

Here is an example using ~I to show
proves ~(B & ~A).


In this derivation line 2 is the assumption
made for the purposes of using ~I.
Since we hope to show "~(B & ~A)" using
~I, we assume "(B & ~A)" and try to derive
an explicit contradiction. After reaching the
explicit contradiction on line 6, we can then
write down on line 7 the negation of the
assumption on line 2. The result on line 7
depends on everything line 6 depends on
except the relevant assumption on line 2.

Notice that just a few changes turns that
derivation into one that shows that
(B & ~A)
proves ~(B→A).


The elimination rule for '~' is almost the same as
the introduction rule. The only difference is that
instead of assuming
ψ and writing down at the end,
you assume
and write down ψ at the end.

~E (Negation Elimination)
If you have assumed ~ψ, and you have derived (φ&~φ),
then you can write down ψ,
depending on everything (φ&~φ) depends on except ~ψ.

As an exercise, here is one example of a
derivation, using ~E. However, the derivation
is incomplete. The names of the rules
are missing. See if you can correctly fill
in the names of rules.

Exercise 3.2a

Show (~A
~B) proves ~(A & B)


You have seen all the rules of our natural
deduction system! Now try making some
derivations on your own.

Exercise 3.2b
Show the following:

~~A proves A

proves (~~A A)

~(P Q) proves (~P & ~Q)