Is the following logically true? ∃x[Cube(x) →∀yCube(y)]

I think that it is logically true. When translated into truth functional form we have: A→B. A truth table shows that it is not a tautology but since one entry in the column is T, it is TT-possible. Thus, logically true.

Any thoughts?


It is an instance of a "logical truth", and specifically an example of universally valid formula of predicate logic, but not a (propositional) tautology.

If all objects are Cube, the formula is True.

The tricky case is when not all objects are Cube: in this case ∀yCube(y) is False.

But if ∀yCube(y) is False, we have some objects a such that Cube(a) is False.

Thus, Cube(a) →∀yCube(y) is False → False, i.e. True.

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You're confusing two different systems of logic. Sentential or propositional logic uses variables to represent whole sentences (hence the name), and connects them with operators such as if-then. Truth tables are used in sentential logic. (First-order) predicate logic extends sentential logic by adding object-predicate constructions (like "x is a cube") and quantifiers ("there is an x such that").

Truth tables can't be used in predicate logic. To see why, consider the sentence (x) x=x (for all x, x is identical with x). This sentence is logically true. But if you tried to represent it in sentential logic, you would get simply p. And p is not logically true — one row of its truth table is false.

(Also, logical truth, logical necessity, and tautology are all synonymous. So in sentential logic, finding a row of the truth table where the sentence is false shows that it's not logically true.)

In more-or-less natural English, your sentence is equivalent to the following: "either something is not a cube or everything is a cube." It will probably be easy to see that this is logically true. (Maura ALLEGRANZA's answer gives more detail here.) But that's not yet a formal proof.

Here's a formal proof using reductio ad absurdum:

WTS: (Ex)[Cx -> (y) Cy]

1. Assume for RAA: -(Ex)[Cx -> (y) Cy]
2. (x) -[Cx -> (y) Cy]
3. -[Ca -> (y) Cy]
4. Ca & -(y) Cy
5. Ca
6. -(y) Cy
7. (Ey) -Cy
8. -Cb
9. -[Cb -> (y) Cy]  (from 2)
10. Cb & -(y) Cy
11. Cb 
-><- (8, 11)
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Here is a proof using the Fitch software (Sorry, it uses P instead of Cube):

enter image description here

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