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In natural language, we often use phrases like:

  1. "fix arbitrary"
  2. "such that"
  3. "we have"

In general, how to translate those phrases into formal logic?

One example condition is:

Fix arbitrary $x,y$. For all $f\in A$ such that $f(x)=0$, we always have $f(y)\geq0$, or we always have $f(y)\leq 0$.

ryang
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High GPA
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1 Answers1

10
  1. "fix arbitrary"

for each

  • Fix an arbitrary $x$ (or:  arbitrarily fix $x$);  then $P(x)$ is true.
    In other words:  $P(a)$ is true; $P(b)$ is true; $P(c)$ is true; etc. $$\forall x\;P(x).$$ Literally:  For each $x,\:P(x)$ is true.
  1. "such that"
  • For each $x$ such that $P(x)$ is true, $Q(x)$ is true. $$\forall x\;\big(P(x)\implies Q(x)\big).$$ Literally:  For each $x,\,$ if $P(x)$ is true, then $Q(x)$ is true.
  • There is some $x$ such that $P(x)$ is true. $$\exists x\;P(x).$$ Literally:  For some $x,\,P(x)$ is true.
  1. "we have"
  • We have $P(x)$ being true.
    (or:  We have that $P(x)$ is true.) $$P(x).$$ More crisply:  $P(x)$ is true.

Fix arbitrary $x,y$. For all $f\in A$ such that $f(x)=0$, we always have $f(y)\geq0$, or we always have $f(y)\leq 0.$

$$∀x\;∀y\;∀f{\in}A\;\Big(f(x)=0\implies f(y)\ge0\;∨\;f(y)\le0 \Big);\tag1$$ more plainly:  for each $f,x$ and $y$ for which $f$ is in $A$ and $f(x)$ equals zero, $f(y)$ is either nonnegative or nonpositive.

Equivalently: $$∀f{\in}A\;\Big(∃x\;f(x)=0\implies ∀y\;f(y)\ge0\;∨\;∀z\;f(z)\le0 \Big),\tag2$$ i.e.,  every function in $A$ that has a root is either identically nonnegative or identically nonpositive,
i.e.,  every function in set $A$ that has a root never crosses (cuts through) the $x$-axis.

(Observe that formalisation (1) is the most succinct, while verbalisation (2) is the easiest to understand.)

ryang
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    This is very concise! In the first formal logic statement, is it possible to have: $g(0)=h(0)=0$ and $g(1)>1$, and $h(1)<1$ simultaneously hold? – High GPA May 28 '22 at 07:35
  • Re: the new plain statement. I guess when we say a function $f$ is nonnegative, it usually mean that $f(\mathbb R)\geq 0$? Will people confuse "$f(y)$ is nonnegative" with "$f$ is nonnegative"? – High GPA May 28 '22 at 07:37
  • Yes. $\quad$ 2. Not technically: $f$ being identically nonnegative means that for each $x$ in $f$'s domain (in real analysis, this is some subset of $\mathbb R),;f(x)≥0.\quad$ 3. "$f(y)$ is nonneg" is an incomplete sentence: the reader is left wondering whether you mean that $f$ is identically nonneg, or that $f$ is nonneg somewhere, and if the latter, whether that nonneg location is unknown or has been specified.
  • – ryang May 28 '22 at 08:28
  • I read your updated statement. However, by saying "$y$ is fixed", I intended to mean that $f(y)\geq 0$ only for the specific $y$ that I fixed, not for any $y$. For example, let's fix $x=0$ and $y=1$. Let $f(0)=g(0)=0$. Then $f(1)\geq 0$ implies $g(1)\geq 0$ because "$f(y)$ is always $\geq 0$ for all $f$ or $f(y)$ is always $\leq 0$ for all $f$". But it is possible that $f(2)\leq 0$. – High GPA May 28 '22 at 10:55
  • In the question I said "Fix $x,y$. Then $f(y)\geq 0$ for all $f$ such that $f(x)=0$, etc..." I was replying to your answer, statement (2). – High GPA May 28 '22 at 11:20
  • Please read paragraph 1 of my just-expanded Answer, then detour to this or this to get a better sense of fixing a variable, arbitrary constants, and of why whether a particular object is being fixed or varying depends on the context. $\quad$ 2. Note that $∀fP(f)∨∀fQ(f)$ isn't equivalent to $∀f\big(P(f)∨Q(f)\big).$
  • – ryang May 28 '22 at 13:33
  • Your example-argument in the third comment above this is invalid: based on your Supplied Proposition and given that $f(0)=g(0)=0,$ we can simultaneously have $f(1)$ being nonneg (thus $f$ being identically nonneg) with $g(1)<0.,$ Verbalisation $(2)$ drops all mentions of $f,x,y,z$ (but not $A)$ as they have merely been placeholders. $\quad$ 4. Statements $(1),&,(2)$ have merely translated your Supplied Proposition; if you had meant to supply Another Proposition, that's a separate Question.
  • – ryang May 28 '22 at 14:01