Question

Show that if \(C\) is a condition that elements of the \(n\)-ary relation \(R\)and \(S\)may satisfy, then \({s_C}(R - S) = {s_C}(R) - {s_C}(S)\).

Short answer

The resultant answer is\({s_C}(R - S) = {s_C}(R) - {s_C}(S)\).

Step-by-step solution

Given data

\({s_C}(R - S) = {s_C}(R) - {s_C}(S)\)is given.

Concept of sets

The concept of set is a very basic one.

It is simple; yet, it suffices as the basis on which all abstract notions in mathematics can be built.\(A\)set is determined by its elements.

If\(A\)is a set, write\(x \in A\)to say that\(x\)is an element of\(A\).

Simplify the expression

The \({s_C}(R - S)\) represents all possible \(n\)-tuples in \(R - S\) that satisfy the condition \(C\)

\({s_C}(R - S) = \{ n{\rm{ - tuple }} \in R - S\mid C\} \)

\({s_C}(R)\)then represents all possible \(n\)-tuples in \(R\) that satisfy the condition \(C\)

\({s_C}(R) = \{ n{\rm{ - tuple }} \in R\mid C\} \)

\({s_C}(S)\)then represents all possible \(n\)-tuples in \(R\) that satisfy the condition \(C\)

\({s_C}(S) = \{ n{\rm{ - tuple }} \in S\mid C\} \)

The difference \(A - B\) of two sets contain all elements in the first set \(A\) that are not in the second set \(B\).

\(\begin{array}{l}{s_C}(R) - {s_C}(S) = \{ n{\rm{ - tuple }} \in R{\rm{ and }}n{\rm{ - tuple }} \notin S\mid C\} \\{s_C}(R) - {s_C}(S) = \{ n{\rm{ - tuple }} \in R - S\mid C\} \end{array}\)

Therefore,\({s_C}(R - S) = \{ n{\rm{ - tuple }} \in R - S\mid C\} = {s_C}(R) \cap {s_C}(S)\).