The king of France -- vacuously speaking

Russell's Theory of Descriptions is still a subject of philosophical discussion, though the theory dates to his Principles of Mathematics published in 1903. His King of France problem, which was posed in Principles, raised for him some issues concerning descriptions.

Yet, that problem, as a matter of strict logic, is easily resolved. I leave the meta-logical issues unaddressed.

Problem:

1. The current king of France is bald.
2. The current king of France is not bald.

How should these two sentences be reconciled, considering that, as France is no monarchy, a statement and its negation both seem to be false?

Solution (not Russell's):

We formulate Russell's puzzle thus:

1. If x is the reigning king of France, x is bald.
2. If x is the reigning king of France, x is not bald.

or, equivalently,

1a. x reigns in France implies x is bald.

1b. x reigns in France implies x is not bald.

It is immediate that both sentences are true under the generally
accepted rules of propositional logic.

'P implies Q' is true when P is false; 'P implies ~Q' also holds when P is false.
That is, a false statement implies anything.

Sometimes statements such as 1. and 2. are called "vacuously true."

Russell, I suppose, was worried about this form:

2. x reigns in France implies x is both bald and not bald.

Hmmm. How can even a nonexistent thing be defined by contradiction? There is no square circle in normal parlance. Nor can there be even a notional king who is simultaneously bald and not bald.

But, this concern evaporates if we put 2. into its equivalent form:

2a. x reigns in France implies x is bald and x reigns in France implies x is not bald.

Item 2a can be expressed

2b. [(P implies Q) & (P implies ~Q)].

The entire statement in braces is true! P does indeed imply both Q and Q's negation. 

2c. P implies (Q & ~Q)

I suppose it is this particular form that prodded Russell to mull over descriptions. Yet, 2c is true. The contradiction (Q & ~Q) is always false, and, as P is false, we have a falsehood implying a falsehood, which yields a truthful implication.

It also should be noticed that the negation ~(P implies Q) is (P doesn't imply Q). The fact that ~Q negates Q is not relevant in this case.

To wit,

3. [(P implies Q) & (P doesn't imply Q)] is always false,

as can be seen by

3a. [(x reigns in France implies x is bald) & (x reigns in France doesn't imply x is bald)]

Similarly,

4. [(P implies ~Q) & (P doesn't imply ~Q)] is also always false,

as is plain in

4a. [(x reigns in France implies x isn't bald) & (x reigns in France doesn't imply x isn't bald)].

Russell, along with the mathematical philosopher A.N. Whitehead helped usher in the era of symbolic, mathematical logic with their Principia Mathematica, published after Russell's Principles.

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Russell, in a discussion of proper names, notes, "Although France is now a republic, I can make statements about the present king of France which, though false, are not meaningless. But if I pretend that he is called Louis XIX, any statement in which 'Louis XIX' is used as a name will be meaningless, not false." The context of that statement was his Great War-era concept of a logic language that could stand in for, say, English. Russell recalls that thinking in My Philosophical Development (Simon and Schuster 1959).

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Russell's problem on the author of Waverley is one of three examples given in his Theory of Descriptions. The Waverley problem even fascinated Godel (see his paper in The Philosophy of Bertrand Russell, Paul Arthur Schilpp ed., Library of Living Philosophers, 1946). The Waverley poser is discussed in a recent Cosmosis post.

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Philosophy Professor Thomas C. Ryckman's excellent page on 3 Russell problems:
https://www2.lawrence.edu/fast/ryckmant/Russell's%20Theory%20of%20Descriptions.htm