**5.** Suppose *x* ∈ *S* − *T*. Then *x* ∈ *S* and *x* ∉ *T*, which means *x* ∈ *S* and *x* ∈ $\overline{T}$, that is *x* ∈ *S* ∩ $\overline{T}$. So *S* − *T* ⊆ *S* ∩ $\overline{T}$. Conversely, if *x* ∈ *S* ∩ $\overline{T}$, then *x* ∈ *S* and *x* ∉ *T*, which means *x* ∈ *S* − *T*, that is *S* ∩ $\overline{T}$ ⊆ *S* − *T*. Therefore *S* − *T* = *S* ∩ $\overline{T}$.

**6.** For Equation (1.2), suppose *x* ∈ $\overline{{S}_{1}\cup {S}_{2}}$. Then *x* ∉ *S*_{1} ∪ *S*_{2}, which means that *x* cannot be in *S*_{1} or in *S*_{2}, that is *x* ∈ ${\overline{S}}_{1}\cap {\overline{S}}_{2}$. So $\overline{{S}_{1}\cup {S}_{2}}$ ⊆ ${\overline{S}}_{1}\cap {\overline{S}}_{2}$. Conversely, if *x* ∈ ${\overline{S}}_{1}\cap {\overline{S}}_{2}$, then *x* is not in *S*_{1} and *x* is not in *S*_{2}, that is, *x* ∈ $\overline{{S}_{1}\cup {S}_{2}}$. So ${\overline{S}}_{1}\cap {\overline{S}}_{2}$ ⊆ $\overline{{S}_{1}\cup {S}_{2}}$. Therefore $\overline{{S}_{1}\cup {S}_{2}}$ = ${\overline{S}}_{1}\cap {\overline{S}}_{2}$.

**12.** (a) reflexivity: Since |*S*_{1}| = |*S*_{1}|, so reflexivity holds.

(b) symmetry: if |*S*_{1}| = |*S*_{2}| then|*S*_{2}| = |*S*_{1}| so symmetry is satisfied.

(c) transitivity: Since ...

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