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Let $f$ be a binary relation between sets $A_0$ and $B_0$ and $g$ be a binary relation between sets $A_1$ and $B_1$.

It can be proved that $[f\cap(A_1\times B_1)=g]\wedge [g\cap(A_0\times B_0)=f]$ implies $f=g$.

Really, it follows $g\cap(A_0\times B_0)\cap(A_1\times B_1)=g$. Thus $g\subseteq A_0\times B_0$. Consequantly $g=f$.

Do the following hold?

  1. $[f\cap(A_1\times B_1)\supseteq g]\wedge [g\cap(A_0\times B_0)]\supseteq f$ implies $f=g$.

  2. $[f\cap(A_1\times B_1)\subseteq g]\wedge [g\cap(A_0\times B_0)]\subseteq f$ implies $f=g$.

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  • $\begingroup$ what exactly is the difference between $\land$ and $\cap$ in this context? $\endgroup$ – Yanko Nov 22 '18 at 17:55
  • $\begingroup$ @Yanko: Their meanings are standard - $\cap$ is intersection of sets, $\wedge$ is conjunction of statements (meaning 'and'). $\endgroup$ – Clive Newstead Nov 22 '18 at 17:56
  • $\begingroup$ @CliveNewstead I get that, but if $f,g$ are binary relations, isn't $f\land g$ the same as $f\cap g$ where $f$ and $g$ are being viewed as subsets of the cartesian product? $\endgroup$ – Yanko Nov 22 '18 at 17:57
  • $\begingroup$ @Yanko: I'd presume so, yes. What's the confusion? $\endgroup$ – Clive Newstead Nov 22 '18 at 17:58
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    $\begingroup$ @Yanko: Ah, you're just parsing it wrong. Read it as $$[f \cap (A_1 \times B_1) = g] \wedge [g \cap (A_0 \times B_0) = f]$$ and so on. $\endgroup$ – Clive Newstead Nov 22 '18 at 18:01
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It is true in general that if $X \subseteq Y \cap Z$, then $X \subseteq Y$ and $X \subseteq Z$. The hypothesis of statement (1) implies that $f \subseteq g$ and $g \subseteq f$, and so $f = g$, as required. So statement (1) is true.

The hypothesis of statement (2) holds for all $f,g$, provided that $A_0 \cap A_1 = \varnothing$ or $B_0 \cap B_1 = \varnothing$, so it doesn't follow in general that $f=g$. So statement (2) is false in general.

In fact, more generally, we have $(A_0 \times B_0) \cap (A_1 \times B_1) = (A_0 \cap A_1) \times (B_0 \cap B_1)$, and so the hypothesis of (2) holds whenever the relations $f$ and $g$ agree on $(A_0 \cap A_1) \times (B_0 \cap B_1)$; but then $f$ and $g$ can say what they like about the elements not in these intersections, so they need not be equal.

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