Suppose that A, B, C and D are sets, R is a relation between A and B

(i.e., R ⊆ A x B), S is a relation between B and C and T is a relation

between C and D. Show the following:

(a) R ¤ (S ¤ T) = (R ¤ S) ¤ T.

(b) (R ¤ S)⁻¹ = S⁻¹ ¤ R⁻¹ .

Please help

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- Sep 19th 2011, 12:09 PMhabsfan31Math Logic: Let A, B, C and D be sets, and R ⊆ A x B, S ⊆ B x C, T ⊆ C x D. Show that
Suppose that A, B, C and D are sets, R is a relation between A and B

(i.e., R ⊆ A x B), S is a relation between B and C and T is a relation

between C and D. Show the following:

(a) R ¤ (S ¤ T) = (R ¤ S) ¤ T.

(b) (R ¤ S)⁻¹ = S⁻¹ ¤ R⁻¹ .

Please help - Sep 19th 2011, 12:27 PMPlatoRe: Math Logic: Let A, B, C and D be sets, and R ⊆ A x B, S ⊆ B x C, T ⊆ C x D. Show
- Sep 19th 2011, 12:33 PMhabsfan31Re: Math Logic: Let A, B, C and D be sets, and R ⊆ A x B, S ⊆ B x C, T ⊆ C x D. Show
Yup, sorry bout that

- Sep 19th 2011, 12:42 PMPlatoRe: Math Logic: Let A, B, C and D be sets, and R ⊆ A x B, S ⊆ B x C, T ⊆ C x D. Show
If think then you have it backwards.

$\displaystyle S\circ R$ exists but $\displaystyle R\circ S$ does not.

Because $\displaystyle R$ relates $\displaystyle A\to B$ and $\displaystyle S$ relates $\displaystyle B\to C$.

So it must be $\displaystyle S\circ R$.

Unless your text material is completely non-standard.

It is usual for compositions to read right to left.

Post Script

Do you understand what I have posted?

If $\displaystyle {\mathcal{G}}\subseteq (U\times V)~\&~ {\mathcal{H}}\subseteq (V\times W) $ then if $\displaystyle U\ne W$ then $\displaystyle \mathcal{H}\circ \mathcal{G}$

exists BUT $\displaystyle \mathcal{G}\circ \mathcal{H}$ may not.

By definition $\displaystyle (u,w)\in \mathcal{H}\circ \mathcal{G} \iff \left( \exists v \right)[(u,v)\in \mathcal{G}~\&~(v,w)\in \mathcal{H}]$.

That means that $\displaystyle \text{Dom}(\mathcal{H})\subseteq\text{Img}(\mathca l{G}).$