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.


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)

Here is a proof using the Fitch software (Sorry, it uses P instead of Cube):

enter image description here

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.