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Subsection 8.4.10 Is It Really Impossible to Prove a Negative?

You’ve probably heard the adage:

It’s impossible to prove a negative.

We now know that that’s ridiculous.

But it certainly is true that some negative claims are indeed very hard to prove.

Consider the claim: There is no life in the universe except on Earth.

To prove this, we’d have to examine the entire universe (and be certain that our examination techniques are foolproof) or we’d have to be able to reason from a sufficiently exact model of the universe that we’d believe our conclusion.

So it’s nearly certain that we’ll never be able to prove this claim.

But some negative claims are straightforwardly provable. This often happens when those claims can be rewritten as positive ones using operators like Quantifier Exchange and De Morgan.

Let’s reconsider the question of whether all birds can fly. Now we’ll assert this specific negative claim:

[1] \(\neg \) \(\forall \)x (Bird(x) \(\rightarrow \) CanFly(x))

This claim is easy to prove. Applying quantifier exchange to [1], we get:

[2] \(\exists \)x \(\neg \)(Bird(x) \(\rightarrow \) CanFly(x))

Applying Conditional Disjunction to [2], we get:

[3] \(\exists \)x \(\neg \)(\(\neg \)Bird(x)  CanFly(x))

Applying De Morgan and Double Negation to [3], we get:

[4] \(\exists \)x (Bird(x) \(\wedge \) \(\neg \)CanFly(x))

To prove [4], it suffices to exhibit our baby emu who cannot fly.

Exercises Exercises

1.

1. Prove or disprove the following negative claim:

[1] ¬∀x (Even(x) → Even(x+1))

Try to prove or disprove [1]. What happened?

  1. It’s possible to prove [1] directly.

  2. It’s possible to prove [1] with a single example (possibly after some logical manipulation of it).

  3. It’s possible to disprove [1] directly.

  4. It’s possible to disprove [1] with a single example (possibly after some logical manipulation of it).

Answer.
Correct answer is B.
Solution.

Explanation. It’s possible to prove [1] with a single example after some logical manipulation:

[1] x (Even(x)  Even(x+1))

[2] x (Even(x)  Even(x+1)) Quantifier Exchange [1]

[3] x (Even(x)  Even(x+1)) Conditional Disjunction [2]

[4] x (Even(x)  Even(x+1)) De Morgan [3]

[5] x (Even(x)  Even(x+1)) Double Negation [4]

Now we prove [5] by exhibiting a single example, say 2. 2 is even. And 2+1 (i.e., 3) is not even.