$H(x,y)$ = "$x$ has $y$", $M(x)$ = $x$ is a man, and $I(x)$ = $x$ has an iPhone. Given these predicates, the English statement that has to be translated into its logical equivalent is "Every man has an iPhone." We can assume the domain is all living people.

Roughly, I interpreted the statement above to be that for every $x$ where $x$ is a man, it would imply that there exists some $y$ where $y$ is an iPhone and the man, or $x$, has the iPhone, $y$. The logical equivalent I came up with is: $$ \forall x (M(x) \implies \exists y(I(y) \ \land \ H(x,y)) $$

Is my interpretation of the English statement correct? Also, just wondering, can the $\exists$ quantifier go with the $\forall$ quantifier in the beginning of the statement; why or why not?


The English sentence is an example of scope ambiguity. There are two readings1:

  1. For every man it holds that he has at least one iPhone (and this might be different iPhones from man to man)
    $\forall x (M(x) \implies \exists y(I(y) \ \land \ H(x,y))$

  2. There is at least one iPhone which every man has (and this is one and the same iPhone, not excluding the possibility that some men might additionally have another iPhone)
    $\exists y I(y)\ \land \ \forall x(M(x) \implies H(x,y))$

Intuitively, your interpreation (1.) is the more plausible one. This is called the linear reading of the sentence, because the surface structure in the sentence corresponds to the underlying logical form (where the univeral quantification has scope over the existential quantification).
If you understand the sentence as "Every man has an iPhone and it is not necessarily one iPhone that everyone has" (which is what you would intuitively understand), then your formalisation is correct.

The other reading is called the inverted reading: Here, the scope relation is the reverse of the one that is reflected in the surface sentence, i.e. the existential quantifier has scope over the unverisal quantifier although it appears later in the sentence.
If you understand the sentence as "There is (at least) one iPhone and this iPhone is owned by every single man", then (2.) is the correct formalisation. This is, in this case, not the most intuitive reading, but possible as well, because the natural language sentence is simply ambiguous. You could imagine other sentences like "Every child sings a song" where it is more understandable to assume that there is a song that is sung by everyone collectively instead of each child singing on its own.

So, yes, both your formalisation (my 1.) and a formalisation with the existential quantifier having scope over the rest (2.) are possible, but they have two different meanings, among which (1.) is more plausible in this context, but since natural language is often ambiguous (surface structure doesn't always reflect underlying logical structurein a 1:1 relationship), the formula shown in (2.) is also possible (but then with a different meaning).

Edit: Revisting your question whether "the ∃ quantifier [can] go with the ∀ quantifier in the beginning of the statement":
Changing the order of the quantifiers does change the logical meaning of the formula. If you move the existential quantifier in front of the univeral one, you get the alternative meaning I formulated in (2.).
Whether you then write $\exists y I(y) \land \forall x (...)$ or $\exists y \forall x I(y) \land (...)$, i.e. whether you move all quantifiers to the very front or place them just at the beginning of their respective scope area is just a matter of style - usually, it is treated the way that the quantifier should not be separated from its restriction, i.e. from where the variables it quantifies over appear in the formula; but the crucial feature that determines the scopal relations is the order of the quantifiers themselves, not the arrangement between the predicates (as long as the quantifier appears before the predicates containing the varible it quantifiers over, otherwise it doesn't make sense anyway), so both of the formulae mentioned in this last paragraph are equivalent in terms of the scopal relationship, as the order between univeral and existential quantifier is the same.

1 My formalisation was based on the assumption that you defined $I(x)$ as "$x$ is an iPhone" (instead of "$x$ has an iPhone") - I don't know whether this was a typo or intended; if you want to treat it as "has an iPhone", then the answer by @Lorenzo is correct and mine is wrong, but it seemed more useful to me to define $I(x)$ as "$x$ is an iPhone" (which I thought you did) so my whole post is based on that definition.


I would do something like this (disregarding the predicates):

Let $M$ be the set of men (why not humans?) and $P$ be the set of iPhone owners.

Then "Every man has an iPhone" could be written as $x \in M \implies x \in P. $

  • $\begingroup$ I understand this approach, but could my solution work? Is it theoretically possible? $\endgroup$ – Shrey Sep 5 '16 at 2:06

Last first: you can use quantifiers every time you need a variable to be quantified and as a consequence even use hundreds of them together. But check some logic basic course to understand all the intricacies of the topic.

Regarding your version you are stating that:

If something (x) is a man then there exists something (y) that has an Iphone and (x) has (y)

Not quite your desired outcome. .

Just cut the $H(x,y)$ predicate and the $y$ variable to obtain the way simpler:

$\forall x (M(x) \implies I(x))$

  • $\begingroup$ I have to use all of the predicates given, so how would I formulate a solution with all of them? Is mine technically wrong? $\endgroup$ – Shrey Sep 5 '16 at 2:12
  • $\begingroup$ @Shrey As you can see above, your solution translates into something that barely makes sense and anyway isn't what you want to achieve. I find it hard to believe that you were given this exact exercise because the predicate $H(x,y)$ requires the variable $y$ to assume the value "Iphone" to denote your english proposition. Unfortunately there's is no way - given these predicates alone - to predicate something like "y is an Iphone": you would need to replace i.e. your $I(x)$ definition with $I(x)=x is an Iphone$. Then your version would work. $\endgroup$ – Lorenzo Sep 5 '16 at 2:53
  • $\begingroup$ I can see where the solution goes south; I said the domain was all living people, but turns out the original problem didn't specify a domain. If I say that the domain is everything (like literally everything), then I(x) would work because x can be an iPhone, right? $\endgroup$ – Shrey Sep 5 '16 at 2:57
  • $\begingroup$ Usually for these propositional exercises you don't need to specify a domain for different types of variables and the common sense "everything" is automatically considered as the domain over which to quantify and predicate. So by simply changing your predicate $I(x)$ you get the job done. If you try to play with domains I'm afraid things would get a little bit messy and take you out of the route these exercises usually follow. This said, yes: if you use only the domain of living people, nothing satisfies predicates H and I (in their "usual" interpretation). $\endgroup$ – Lorenzo Sep 5 '16 at 3:16
  • $\begingroup$ Your formlation doesn't seem to make any sense to me. What you said is basically "Every x which is a man is an iPhone". This is clearly not what you want. 1) You can not con-indice (use the same variable for) the man and the iPhone, 2) you need the H(x,y) as a predicate for the "has"-relation and 3) you need an existential quantification due to the "an iPhone" part. What your formlua says doesn't reflect the orignal meaning at all. $\endgroup$ – lemontree Sep 5 '16 at 13:38

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