Does this statement reduce to True or False? How would this statement be said in words?

$(y)((x)Fx\supset Fy)$

What is the thought that those symbols are trying to communicate to me?

The book this question comes from was published in 1963 by Harvard U. The second page of the first chapter. Are there any older Ivy grads out there?

  • 2
    $\begingroup$ I suppose this notation means what most would rather write as $\forall y\forall x(Fx\to Fy)$. But what is $F$? $\endgroup$ Sep 17, 2016 at 13:42
  • $\begingroup$ @Ivan Hieno: For all $y$ and for all $x$ it is a fact that if $Fx$ holds, then so does $Fy$. The notation of Hagen von Eitzen's comment seems correct to me. Of course, it could be possible to also write $\forall y: ((\forall x: Fx) \to Fy)$ which is not the same. $\endgroup$
    – Moritz
    Sep 17, 2016 at 13:47
  • 1
    $\begingroup$ The second interpretation, $\forall y((\forall xFx)\to Fy)$, is presumably the correct one, because then the formula is true. $\endgroup$
    – TonyK
    Sep 17, 2016 at 16:34

2 Answers 2


In this notation, $(x)$ means "for all $x\!",$ and $\supset$ means "implies". $F$ is a symbol for a one-place predicate (or relation).

From the placement of the parentheses, I would interpret this to mean:

For every $y,$ the following is true: If $Fx$ is true for all $x,$ then $Fy$ is true.

(And, as you can see, this statement is true.)

To be sure of this interpretation, you'd need to look up the author's recursive definition of formula, which would implicitly tell you the order of precedence here between quantification and the connective $\supset.$

Added: I found the reference -- it's Set Theory and Its Logic, by Quine. It appears that he doesn't really give a detailed recursive definition of formula, but the mention of $(y)((x)Fx\supset Fy)$ says that this is valid, so the interpretation I suggested is correct.


I think that the statement can be resolved into a boolean value, then we have four possibilities.

Case 1

y = {2,4,6}; x = {10,12,14}; F = "is even" then

$(y)((x)Fx\supset Fy)$ = true

Case 2

y = {2,4,7}; x = {10,12,14}; F = "is even" then

$(y)((x)Fx\supset Fy)$ = false

Case 3

y = {2,4,6}; x = {10,12,15}; F = "is even" then

$(y)((x)Fx\supset Fy)$ = true

Case 4

y = {2,4,7}; x = {10,12,15}; F = "is even" then

$(y)((x)Fx\supset Fy)$ = true


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