I have the sentence "If a cube is to the right of a dodecahedron but not in the back of it then it is as large as the dodecahedron." I have to translate this sentence into logical symbols. I can translate a similar sentence that says "if a cube is to the right of a dodecahedron but not in the back of it then it is large"by saying ∀x ((cube(x) ∧ ∃y (dodec(y) ∧ rightof(x, y) ∧ ¬backof(x, y)) → (larger(x))). Here it says for all objects (x), if the object is a cube and there exists at least one object (y) that is a dodecahedron, the cube is to the right of it, and it's not in the back of it then the cube is large. The problem with the first sentence is that at the end of the translation we've lost our dodec after the arrow so we can't just reference y again. Here's one way I've tried to translate the first sentence:

∀x (Cube(x) → ∃y (Dodec(y) ∧ RightOf(x, y) ∧ ¬BackOf(x, y) ∧ (Larger(x, y) ∨ SameSize(x, y)))) where to represent the phrase "as large as" I'm saying it's either larger or the same size. I had to do some confusing things with this translation so I'm not quite sure if this is correct.


1 Answer 1


You can think of your claim as making a claim about a pair of objects: if one is a cube and the other is a dodecahedron, and they are related in such-and-such a way, then they are related in this-and-that way. So, you get:

$$\forall x \forall y ((Cube(x) \land Dodec(y) \land RightOf(x,y) \land \neg BackOf(x,y)) \rightarrow Samesize(x,y))$$

(BTW: I took 'as large as' as having the same size ... otherwise I think it would have said 'at least as large as')

If you insist on having the second quantifier inside:

$$\forall x (Cube(x) \rightarrow \forall y ((Dodec(y) \land RightOf(x,y) \land \neg BackOf(x,y)) \rightarrow Samesize(x,y)))$$

So the trick here is really that that second quantifier is also a universal ... just as 'a cube' is used in the sentence as 'any cube' and thus translates into a universal, the 'a dodecahedron' in this sentence also means 'any dodecahedron' !

OK, you ask, so how come we got an existential in that other sentence?

Well, please note that if $Q$ does not contain any $x$ as a free variable, then:

$$\exists x \ P(x) \rightarrow Q$$

is equivalent to:

$$\forall x \ (P(x) \rightarrow Q)$$

And so likewise your:

$$\forall x (Cube(x) \rightarrow (\exists y (Dodec(y) \land RightOf(x,y) \land \neg BackOf(x,y)) \rightarrow Large(x)))$$

is equivalent to:

$$\forall x (Cube(x) \rightarrow \forall y ((Dodec(y) \land RightOf(x,y) \land \neg BackOf(x,y)) \rightarrow Large(x)))$$

And if you compare that with my second translation of the first sentence above you now see how the two sentences really do have the same form!

OK, but then why can't we likewise use an existential for the first sentence? Well, that's because we can't use the earlier equivalence, and that's because the $SameSize(x,y)$ has $y$ as a free variable.

  • $\begingroup$ Ok so my problem was that I thought I had to use an existential because it said "a dodec." $\endgroup$
    – Josh Susa
    Commented Nov 16, 2017 at 2:07
  • $\begingroup$ @JoshSusa Yes, that was it! $\endgroup$
    – Bram28
    Commented Nov 16, 2017 at 2:08

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