# Simple predicate calculus

All I want to do is write the following things into notation.

My trouble is in inserting the correct order and understand where to use the same variables.

The predicates are Students, Answers, and Questions.

1. None of the students answered all of the questions

2. At least one student answered all of the questions

3. One student answered all of the questions.

I get confused in the differences in how to show one student vs at least one

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The predicates are Students, Answers, and Questions.

So, you need two variables: say $x$ and $y$ for all but the 3rd translation:

Let $Sx$ denote "$x$ is a Student";

Let $Axy$ denote "$x$ answered $y$";

Let $Qy$ denote "$y$ is a question."

In each case you are going to need the existential quantifier to assert or deny the existence of a student and the universal quantifier to convey that we are making a statement about all questions.

$(1)$ None of the students answered all of the questions.

$(1)$ is the denial of existence. We can keep both quantifiers needed for this statement at the beginning by asserting the equivalent of $(1)$:

There does not exist an x such that for all y [if (x is a student and y is a question) then (Axy)].

Can you try working with translating that using symbolic logic?

• It is helpful, sometimes, to use "pseudo logic" as an intermediate step going from english to logic, or vice versa. Pay special attention to the parentheses.

$(2)$ At least one student answered all of the questions.

For $(2)$ we assert the existence of a student (i.e. "there exists at least one $x$ such that for all y, (if x is a student and y is a question, then....

$(3)$ One student answered all of the questions.

Here you need to assert that there exists one and only one student such that...

If you have studied the "existence-of-a-unique" quantifier $\exists !$, then your task is easy: you need only replace on quantifier used in $2$
If not, we can still express uniqueness using only the symbols $\exists, \;\;\forall,\;\; \text{and}\;\;=.$ We want to

Here you can still use your translation of $(2)$ - but you need to qualify it with "for all z (if z is a student and z answered all the questions, then z must be x)."

See if the following makes sense:

$(3)\;\;\exists x \forall y\{[(Sx \land Qy) \rightarrow Axy] \land \forall z [(Sz \land Azy) \rightarrow (z=x)]\}.$

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That should be identity and not negation in the last sentence. – Peter Smith Nov 6 '12 at 21:08
@Peter: I was including the symbols needed to assert uniqueness barring the use of $\exists !$. So negation will help. But yes, I need to add the equality/identity too...thanks! – amWhy Nov 6 '12 at 21:13
@Peter You're right about negation...not needed here...one could, I suppose, end with "$\forall z$ [(if $Sz \land z \neq x)$, then...$\exists y(Qy \land \lnot Azy)]$, but that would be, I suspect, a bit messier! – amWhy Nov 6 '12 at 23:02
@amWhy: Another long answer. Sweet +) – Babak S. Aug 10 '13 at 11:13

at least one is translated as there exists $\exists$ Existential quantification

none is translated as does not exist $\not\exists$ denial of quantification

one is translated as exactly one exist $\exists_1$ or $\exists !$ uniqueness quantification

all is translated as for all $\forall$ Universal quantification

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I have recommended it before here: but if you are puzzled about simple cases of rendering propositions into logical notation, then try reading the excellent opening chapters of vol. 2 of Paul Teller's Modern Formal Logic Primer, available freely at http://tellerprimer.ucdavis.edu

But note that if you want to translate 'One $F$ is $G$' -- meaning exactly one $F$ is $G$ -- then you need to have the identity predicate available in your logical notation.

You can render 'Exactly one $F$ is $G$' as 'At least one $F$ is $G$ $\land$ at most one $F$ is $G$'. The second clause is in turn rendered

$$\forall x\forall y ([(Fx \land Gx) \land (Fy \land Gy)] \to x = y)$$

Or you can render 'Exactly one $F$ is $G$' as 'At least one $F$ is $G$ $\land$ any $F$ which is $G$ is that same thing',

$$\exists x((Fx \land Gx) \land \forall y((Fy \land Gy) \to y =x)$$

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Teller's Primer is a great resource! And...free is good! It's one of the two primary texts used at the University I teach at for a second-semester undergraduate course in logic, the other being Mates Elementary Logic, 2nd edition, Oxford University Press, 1972. – amWhy Nov 8 '12 at 1:20