Proof Strategies
A Supplement to Chapter 19
When you are constructing proofs for the first time, it is natural to ask: Where do I start? How do
I know which steps to take? How do I begin a proof? Unfortunately, there is no useful algorithm
or decision procedure that you can use to answer these questions. In short, there is no practical,
mechanical procedure for constructing a proof. Building a proof is therefore somewhat of an art.
Without a practical decision procedure that tells you how to construct a proof, you must proceed
by a combination of trial and error, intuition, pattern recognition, general problem-solving skills,
and a strategy we may call “working backward from the conclusion” Although these strategies
are not decision procedures, they may help figure out where to begin and they may help as you
construct proofs.
THE TRIAL AND ERROR STRATEGY
Suppose you begin a proof and are unsure where to start or where to go. Scan the premises,
looking for any of the inference rule patterns. When you spot a pattern, apply the appropriate rule
and make the required move, whether or not it seems to take you to the conclusion. Often, if you
make enough moves like this, even though you have no overall plan or direction in mind, you
will eventually arrive at the conclusion. Let us call this the trial and error strategy. For example,
consider the following argument:
1. (A v E) > ~ G
2. G v (R & H)
3. (A v E) / H
(Note that we are using the greater than sign, >, for the horseshoe.) When you have become
familiar with the inference rule patterns, you should be able to spot the MP pattern on lines 1 and
3. This gives you
1. (A v E) > ~G
2. G v (R & H)
3. A v E / H
4.
~G
MP 1, 3
Next, look for another pattern. Notice that lines 2 and 4 instantiate the premise section of DS.
Thus:
1. (A v E) > G
2. G v (R & H)
3. A v E / H
4.
~G
MP 1, 3
5. R & H
DS 2, 4
Finally, notice that line 5 fits the premise section of Simp, which allows you to infer the
conclusion:
6. H
Simp 5
If you are proceeding by the trial and error strategy, the most important piece of advice is this: If
you spot a pattern and see an inference you could make, but you don’t see where it would take
you, don’t hesitate . . . just do it! When you infer another line, another pattern will probably open
up, which will probably lead to another inference, and another, until you eventually reach the
conclusion.
THE DECOMPOSITION STRATEGY
Another proof strategy may be called the decomposition strategy: Break larger sentences down
into smaller sentences by deriving parts of the larger sentences. Use these smaller parts to derive
additional lines, until you finally derive the conclusion. For example, consider the following
symbolized argument:
1. (H v C) > [(E v F) v (A & B)]
2.
~(E v F)
3. H / B
Let us begin by deriving something that we can use to break up (decompose) line 1:
4. H v C
Add 3
Using this, plus MP, we can now bring down the consequent of 1:
5. (E v F) v (A & B)
MP 1, 4
Notice that line 2 is the negation of the left disjunct of line 5. Consequently, we can use DS to
break up 5 as follows:
6. A & B
DS 2, 5
The last move is obvious:
7. B
Simp 6
Notice that in each step after line 4 we broke a larger sentence down by deriving one of its parts.
WORKING BACKWARD FROM THE CONCLUSION
Let’s suppose we are trying to prove the following symbolized argument valid:
1. ~H > (B > A)
2. S v B
3. H > E
4. ~S
5. ~ E
/A
Ultimately, we want to reach--at the bottom of the proof--the conclusion A. Let’s begin with the
conclusion and trace a series of steps backward to the first step in the proof. First, where in the
premises is the conclusion formula? Notice that A is at the end of line 1. How might we derive
it? On line 1 the sentence A is the consequent of a conditional. What rule would let us derive the
consequent of a conditional? Modus Ponens. What would MP require? First, the conditional
must be all by itself on a line. Second, another line must consist of just the antecedent of the
conditional. So, first we will need to bring the conditional B > A down onto a line by itself, and
then we will need a line consisting of B by itself. From those two lines, A will follow. The
inference would look like this:
B>A
B
Therefore:
A
How might we “detach” or bring down B > A? Notice that on the first line,
B > A is itself the consequent of a conditional. If we could get the antecedent of that conditional
on a line by itself, we could use MP to bring down B > A. The inference would look like this:
~H > (B > A)
~H
Therefore,
B>A
But how might we get ~H on a line by itself? First, find H in the premises. In line 3, H is the
antecedent. Modus Tollens would let us infer ~H from line 3, if we also had on another line the
negation of the consequent of line 3. The inference would look like this:
H>E
~E
Therefore,
~H
So we need ~ E on a line by itself. How will we get that? We locate ~E on line 5. Next, we need
to derive B in order to apply MP to B > A. How will we get B? Notice that B is the right disjunct
of the disjunction on line 2. If we can derive or find, on a line by itself, the negation of the left
disjunct of that disjunction, we can infer B, which we may then use to bring down our
conclusion, A. The negation of the left disjunct of line 2 is ~S. So, we will need ~S in order to
break B out of line 2. Looking around, we find ~S on line 4. Putting this together in the proper
order, we get:
1. ~H > (B > A)
2. S v B
3. H > E
4. ~S
5.
~E / A
6. ~H
MT 3, 5
7. B > A
MP 1, 6
8. B
DS 2, 4
9. A
MP 7, 8
So, approach a proof the way you would a game of strategy such as chess or Monopoly.
Constantly think about the various possibilities open to you at each step. Look ahead to where
you want to be, and think of ways to get there from where you are. In a game of strategy, a
skilled player develops the ability to look several steps ahead and plans moves in terms of that
goal. A skilled logician does the same thing. You can do it!
End