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Consider this set of propositions:

P1. The sentence “The current king of France is bald.” expresses a proposition.

P2. The sentence “The current king of France is not bald.” expresses a proposition.

P3. The propositions expressed by the sentences “The current king of France is bald.”and “The current king of France is not bald.” are contradictory.

P4. For all p and for all q, if p is a contradictory of q, then (p is true and q is false, or q is true and p is false).¹

P5. Iff the sentence “The current king of France is bald.” expresses a proposition, then the sentence “The current king of France is not bald.” expresses a proposition.

P6. If the sentences “The current king of France is bald.” and “The current king of France is not bald.” express propositions, then the propositions are both false.

These propositions form an inconsistent set. When a person is confronted with that he believes is inconsistent, then that person will almost always stop believing at least one of the things he believes. However in this case people usually stop believing both (P1) and (P2), and that is because of (P5). To stop believing one of them only would be futile as the set would still be inconsistent. I suppose that the same people find the other propositions in the set much more justified than (P1) and (P2), and that is the reason why they are rejected.

About (P3)

But why is it that people believe (P3)? It seems to me to be a interpretation failure. Perhaps they interpret the two sentences:

S1. The current king of France is bald.

S2. The current king of France is not bald.

As:

P7. (∃x)(Kx∧Bx)

P8. ¬(∃x)(Kx∧Bx)

(P7) and (P8) may be read as:

P7′. There exists at least one x such that he is the current king of France and he is bald.

P8′. It is not the case that there exists at least one x such that he is the current king of France and he is bald.

(P7) and (P8) are rightly seen as contradictory, but they are not equivalent to the propositions expressed by (S1) and (S2). To see this notice where the negation in placed in (S2). It is not placed in the front of the sentence as it is in (P8), and neither is it placed in the first sentence part.² It is placed in the second sentence part, and it is that part that is negated, not the entire sentence. The correct interpretation of the propositions expressed by (S1) and (S2) is therefore:

P9. (∃x)(Kx∧Bx)

P10. (∃x)(Kx∧¬Bx)

They may be read as:

P9′. There exists at least one x such that he is the current king of France and he is bald.

P10′. There exists at least one x such that he is the current king of France and he is not bald.

Clearly, now we can see that the propositions expressed by (S1) and (S2) are not contradictory but they are merely contrary, that is, they cannot both be true but they can both be false. This implies that (P3) is false, and so the threat to the set is removed.³

Why (P6) is true

Additionally we can now see why (P6) is true. Before we just thought (P6) evidenced by our intuitions. But since both (P9) and (P10) imply (by simplification) that:

P11. (∃x)(Kx)

P11′. There exists at least one current king of France.

And, as we know there is no current king of France, therefore, (P11) is false. If (P11) is false, then so are (P9) and (P10). If they are both false, then (P6) is true.

We can also generalize a bit from this. What seems to confuse people about (S1) and (S2) is that the subject, that is, the current king of France, is missing, that is, he does not exist.

Appendix

Here is a predicate logic version of the original set.

Translation key

Ex ≡ x expresses a proposition.

C(AB) ≡ A and B are contradictories.

a ≡ (The sentence) “The current king of France is bald.”

b ≡ (The sentence) “The current king of France is not bald.”

A ≡ (The proposition) “The current king of France is bald.”

B ≡ (The proposition) “The current king of France is not bald.”

The set

P1′. Ea.

P2′. Eb.

P3′. (Ea∧Eb)→C(AB)

P4′. (∀P)(∀Q)(C[PQ]→[(P∧¬Q)∨(Q∧¬P)])

P5′. Ea↔Eb

P6′. (Ea∧Eb)→(¬A∧¬B)

Notes