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Subsection 8.2.4 Don’t Do Proofs Backwards

A common error, in attempting to construct a direct proof, is to start from the conclusion and work backwards, rather than to start from the premises and work forwards. While it often helps to search for a proof by working backwards from the conclusion, a valid proof must proceed from premises to conclusion, not the other way.

We can see how going backwards can fail to produce a legitimate proof with a very simple example from Boolean logic. Give names to the following statements:

C: Cody wins the prize.

J: Jody wins the prize.

P: There will be a party.

Assume the following premises:

[1] C  P If Cody doesn’t win the prize, there will be no party.

[2] P There will be no party.

We’d like to prove:

[3] (C  J) Neither Cody nor Jody wins the prize.

Suppose that we try to prove [3] by starting with it and reasoning backwards to the premises:

[3] (C  J) Goal

[4] C  J De Morgan [3]

[5] C Simplification [4]

[2] P Modus Ponens [1], [5]

So we’ve derived something that we know must be true since it’s one of the premises.

But this isn’t a proof. If our conclusion is true, so are our premises. But there’s no new information there. The premises are true, regardless of the truth status of the conclusion.

But maybe, since we’ve found a chain of reasoning that connects the conclusion to the premises (albeit backwards), we can transform what we’ve written into a valid proof. Let’s start from the premises and go the other way. What happens?

[2] P Premise

We’re stuck right away. We can’t apply Modus Ponens backwards to derive C. And by the way, if we had made it past this step, the reasoning that derived [5] also cannot be reversed. If we know P  Q, we can use Simplification to derive P. But, based just on P, we cannot prove anything about Q, and in particular we cannot prove P  Q. So this “proof” used two irreversible rules.

We shouldn’t be surprised that we can’t reverse this proof. From our premises, it isn’t possible to conclude anything about either Cody or Jody winning the prize, much less some stronger claim about both of them.

The problem we just saw is that, while logical identities can be used in either direction, there are inference rules that are one way streets. If we try to use them backwards they will take us right into proof Never Never Land.

If, on the other hand, we have used only reversible rules, then it’s possible to convert an invalid, backwards “proof” into a valid actual proof simply by reversing it. Remember, though, that if you’re tempted to try to find a proof by working backwards and you’re lucky enough to find a reversible one, you must actually do the reversal before claiming that you have a proof.

Prove that, for every positive real number x, x + 4/x  4.

Proposed “proof”:

[1] Let x be a positive real number and suppose that x + 4/x  4.

[2] Multiplying by x, we get x2 + 4  4x.

[3] Subtracting 4x from both sides, we get x2 - 4x + 4  0.

[4] Factoring that, we get (x - 2)2  0.

Now we observe that [4] must be true, since the square of any real number is nonnegative.

The problem here is that what we have is a proof of the following: If it’s true that, for every positive real number x, x + 4/x  4, then some other thing [4] is also true. But that other thing is always true, independent of the truth of the claim we’re interested in. So our proof hasn’t actually yielded anything new. Most importantly, what we don’t have is a proof of our claim. We assumed it was true; we didn’t prove it true.

But we have discovered a chain of reasoning between something that is already known to be true [4] and the claim that we want to prove [1]. And, in this case, every step that we performed is, in fact, reversible. So here’s a valid proof of our claim:

[1] Since the square of every real number is nonnegative, we have, in particular:

(x - 2)2  0.

[2] Expanding that, we have x2 - 4x + 4  0.

[3] Adding 4x to both sides, we have x2 + 4  4x.

[3] Assume that x is positive. Then we can divide by it, which yields x + 4/x  4.

[4] Thus, for every positive real number x, x + 4/x  4.

Notice that the assumption that we need in order to carry out step [3] becomes a condition (a guard) on the claim that we can ultimately make.

Exercises Exercises

1.

Assume the following premises:

[1] P

[2] PR

[3] R → ¬Q

We want to prove: ¬(PQ)

Consider the following proposed proof:

[4] ¬(PQ)

[5] ¬(¬PQ) Conditional Disjunction [4]

[6] P ∧ ¬Q De Morgan [5]

[1] P Simplification [6]

Which of the following statements is true:

  1. This proof is correct.

  2. This proof isn’t correct. But the claim is true and it’s possible to write a correct proof by reversing the order of the steps of this one.

  3. This proof isn’t correct. But the claim is true and it’s possible to write a correct proof by reversing the order of some of the steps of this one and then adding additional steps

  4. This proof isn’t correct although the claim is true. But nothing in this proof will help us find a correct proof since no steps are reversible.

  5. This proof isn’t correct and the claim isn’t true.

Answer.

Correct answer is C

Solution.

Explanation: This proof is wrong because it starts by assuming the claim that is to be proved. However, the step from [4] to [5] and the step from [5] to [6] are both reversible. And we can use those reverse steps but we need something else first. Here’s a correct proof:

[1] P Premise

[2] P  R Premise

[3] R  Q Premise

[4] R Modus Ponens [1], [2]

[5] Q Modus Ponens [3], [4]

[6] P  Q Conjunction [1], [5]

And now we can reverse the first two steps of the incorrect proof to get:

[7] (P  Q) De Morgan [6]

[8] (P  Q) Conditional Disjunction [7]

So going backward initially, while it didn’t produce a correct proof, pointed us to an intermediate result ([6]) that we could use. So we constructed a correct proof of it and then moved forward to the real conclusion.

2.

Consider the claim that, for any nonnegative reals a and b:

[1] \(\frac{a + b}{2}\)\(\sqrt{ab}\)

Consider the following proposed proof of [1]:

[1] \(\frac{a + b}{2}\)\(\sqrt{ab}\) Squaring both sides, we get:

[2] \(\left( \frac{a + b}{2} \right)^{2}\)ab

[3] \(\frac{{(a + b)}^{2}}{4}\)ab

[4] (a + b)2 ≥ 4ab

[5] a2 + 2ab + b2 ≥ 4ab

[6] a2 - 2ab + b2 ≥ 0

[7] (ab)2 ≥ 0

[7] must be true since the square of any real number is nonnegative. Q. E. D.

Which of the following statements is true:

  1. This proof is correct.

  2. This proof isn’t correct. But the claim is true and it’s possible to write a correct proof by reversing the order of the steps of this one.

  3. This proof isn’t correct although the claim is true. But it isn’t possible to write a correct proof just be reversing the order of the steps of this one. There is exactly one irreversible step.

  4. This proof isn’t correct although the claim is true. But it isn’t possible to write a correct proof just be reversing the order of the steps of this one. There are at least two irreversible steps.

  5. This proof isn’t correct and the claim isn’t true.

Answer.
Correct answer is B.
Solution.
Explanation: This claim is correct. It even has a name. It’s called the Arithmetic Mean/Geometric Mean Inequality. The proof is wrong because it’s backwards. But every step is reversible. (Note that often squaring isn’t reversible, but in this case it is because we are guaranteed that both a and b are nonnegative.)