The logical equivalence of two statements

Question

Suppose that $(R,\prec)$ is a totally ordered set, and let $S\subseteq R.$ Define: $$P:=\forall x\in S \ \forall y\in R(y\preceq x\Rightarrow y\in S),$$$$Q:=\forall x\in S (\forall y\in R(y\preceq x)\Rightarrow y\in S).$$

Are $P$ and $Q$ logically equivalent? I believe they are. However, their negations $$\neg P = \exists x\in S \ \exists y\in R(y\preceq x\land y\notin S),$$ $$\neg Q=\exists x\in S (\forall y\in R(y\preceq x)\land y\notin S)$$ do not appear to be logically equivalent. Where is my mistake?

Answer 1

$$P:=\forall x{\in} S \ \forall \color{cyan}y{\in} R\ (\color{cyan}y\preceq x\Rightarrow \color{cyan}y\in S)\tag1$$ $$Q:=\forall x{\in} S\ (\,\forall \color{green}y{\in} R\ (\color{green}y\preceq x)\Rightarrow \color\red y\in S\,)\tag2$$

$Q,$ but not $P,$ contains a free occurrence of $\color\red y$ (by convention, the scope of the green $\color{green}y$ is actually just ∀y∈R y⪯x). So, $P$ can't possibly be logically equivalent to $Q.$

$$\neg P = \exists x{\in} S \ \exists \color{cyan}y{\in} R\ (\color{cyan}y\preceq x\,\land\, \color{cyan}y\notin S)\tag3$$ $$\neg Q=\exists x{\in} S\ (\ \forall \color{green}y{\in} R\ (\color{green}y\preceq x)\,\land\, \color{red}y\notin S\ )\tag4$$

Yes, you've negated $P$ and $Q$ correctly.

Are $P$ and $Q$ logically equivalent?

Now, \begin{align}Q&\equiv\forall x{\in} S \,\big((\forall \color\green z{\in} R\;\color\green z\preceq x)\Rightarrow \color\red y\in S\big)\tag2\\&\equiv \forall x{\in} S \,\exists \color{green}z{\in} R\,\big(\color{green}z\preceq x\Rightarrow \color\red y\in S\big)\tag{2A}\end{align} \begin{align}\lnot Q&\equiv\exists x{\in} S \,\big((\forall \color\green z{\in} R\;\color\green z\preceq x)\,\land\, \color\red y\not\in S\big)\tag4\\&\equiv\exists x{\in} S \,\forall \color{green}z{\in} R\,\big(\color{green}z\preceq x\,\land\, \color\red y\not\in S\big)\tag{4A}\end{align} (the second and fourth logical equivalences (≡) are explained in https://en.wikipedia.org/wiki/Prenex_normal_form#Conversion_to_prenex_form ).

Comparing (1) and (2A), and separately comparing (3) and (4A), confirms that $$P\not\equiv Q\\ \lnot P\not\equiv \lnot Q.$$

Where is my mistake?

Falsely believing that $(\forall y{\in} R\;y\preceq x)\Rightarrow y\in S\tag*{}$ is logically equivalent to $\forall y{\in} R\,\big(y\preceq x\Rightarrow y\in S\big)\tag*{}$ misled you, at the start, to think that $P$ is equivalent to $Q.$

Suppose that $(R,\prec)$ is a totally ordered set, and let $S\subseteq R.$

The logical equivalence of $P$ and $Q$ is actually indepdendent of this mathematics.

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Reply to the OP's comment

Could you please explain why you're using a different color for y, and why you replaced y with z?

Sure. The green $\color{green}y,$ the cyan $\color{cyan}y$ and $x$ are distinct dummy/bound variables, so they can be renamed (though for each colour, any renaming must apply to every instance of the variable) to clarify this (instead of recycling the variables); in contrast, the red $\color{red}y$ is a free variable, so cannot be renamed. Analogously, $$\int_0^{\color{red}y} t\, \mathrm dt=\int_0^{\color{red}y} u\, \mathrm du\ne\int_0^x u\, \mathrm du.$$

Answer 2

In the statement $Q$, you cannot restrict the quantifier $\forall y\in R$ to the formula $y\preceq x$ only, since $y$ appears in the formula $y\in S$. So it doesn't make sense.

Some more explanations. You can have a free variable in a formula, or you can bind it with a quantifier. For instance,

• $x\preceq y$ is a relation between two elements ;
• $\forall x (x\preceq y)$ means that $y$ is a maximum ;
• $\forall x,\forall y, x\preceq y$ means that your set cannot have two distinct element ;
• $\exists x,\forall y:x\preceq y$ means that your set has a minimum element ;
• $\exists x,\exists y:x\preceq y$ means only that your set is nonempty ;
• $\forall x,\exists y:x\preceq y$ is always true for a reflexive relation.

But you cannot use the same variable as free and binded in the same formula, otherwise you would not know if it concerns the binded or the free variable.