1) $R,S,T$ are relations on the same set.
Prove that $R(S\cup T)=RS\cup ST$
The proof that I stumbled upon was the following:
$(a,b)\in R(S\cup T)⇒((a,x)\in R)∧((x,b)\in S∨(x,b)\in T)⇒(a,b)\in RS∨(a,b)\in RT$
At this stage I got stuck. How exactly does the last transition make sense? Could someone please explain this?
2) When multiplying relations, let's say the relations $R$ and $S$ - do we say that $(x,y)\in R ∨ (x,y)\in S$ ?
Or perhaps $(x,y)\in R ∧ (x,y)\in S$? ('and' instead of 'or')
And why is it?
Thank you.
$(a,b) \in RZ \iff (a,x)\in R \land (x,b) \in Z$
This means: $a$ and $b$ are related by the product $RZ$ if, and only if, there exists an $x$ such that $a$ and $x$ are related by $R$, and $x$ and $b$ are related by $Z$.
$(a,b) \in S\cup T \iff (a,b) \in S \lor (a,b) \in T$
This means: $a$ and $b$ are related by the union of $S$ and $T$ if, and only if, either $a$ and $b$ are related by $S$ or $a$ and $b$ are related by $T$.
Put it together.
$$(a,b) \in R(S\cup T) \\ \iff (a,x)\in R \land (x,b) \in (S\cup T) \\ \iff (a,x)\in R \land \Bigl((x,b) \in S \lor (x,b) \in T\Bigr) \\ \iff \Bigl((a,x)\in R\land (x,b) \in S\Bigr)\lor\Bigl((a,x)\in R \land (x,b) \in T\Bigr) \\ \iff (a,b) \in RS \lor (a,b)\in RT \\ \iff (a,b)\in RS\cup RT \\\therefore R(S\cup T)\equiv RS\cup RT$$