Specialization and Generalization

Here, we are going to implore a different style of proof. In all previous proofs we had an expression of the form $\text{LHS} = \text{RHS}$. We then start with the $\text{LHS}$ at the top and arrive at the $\text{RHS}$ at the bottom. In these expressions, there is are equals sign. Instead, we are going to start with the entire expression at the top and arrive at $\color{green}T$ at the bottom. Since the original statement is equivalent to $\color{green}T$, it must be a true expression. We are essentially using the identity law

\[a = \color{green}T \quad = \quad a\]

Similar to Absorption, these relations may seem simple and obscure, but they turn out to be very useful in practice.


Specialization

$(a \wedge b) \Rightarrow a$

\[\begin{align} &(a \wedge b) \Rightarrow a \\ &= (\overline{a \wedge b}) \vee a &&\text{Material Implication} \\ &= (\overline{a \wedge b}) \vee (a \wedge \color{green}T) &&\text{Identity} \\ &= (\overline{a \wedge b}) \vee (a \wedge (b \vee \overline{b})) &&\text{Excluded Middle} \\ &= (\overline{a \wedge b}) \vee ((a \wedge b) \vee (a \wedge \overline{b})) &&\text{Distributivity} \\ &= ((\overline{a \wedge b}) \vee (a \wedge b)) \vee (a \wedge \overline{b}) &&\text{Associativity} \\ &= \color{green}T \vee (a \wedge \overline{b}) &&\text{Associativity} \\ &= \color{green}T &&\text{Base} \end{align}\]


Generalization

$a \Rightarrow (a \vee b)$

\[\begin{align} &a \Rightarrow (a \vee b) \\ &= \overline{a} \vee (a \vee b) &&\text{Material Implication} \\ &= (\overline{a} \vee a) \vee b &&\text{Associativity} \\ &= \color{green}T \vee b &&\text{Excluded Middle} \\ &= \color{green}T &&\text{Base} \end{align}\]

Boolean Algebra Series