Match the following list :

List – I (Boolean logic function) 

List – II (Inverse of function) 
(a) 
ab + bc + ca 
(i) 
a̅ (b̅ + c̅) 
(b) 
ab + a̅ b̅ + c̅ 
(ii) 
a̅ b̅ + b̅ c̅ + c̅ a̅ 
(c) 
a + bc 
(iii) 
(a ⊕ b) c 
(d) 
(a̅ + b̅ + c̅) (a + b̅ + c̅) (a̅ + b̅ + c) 
(iv) 
abc + a̅bc + abc̅ 
Concept:
The inverse of the function is nothing but the complement of the given expression.
Also, DeMorgans Theorem states that:
\(\left( i \right)\;\overline {{A_1} \cdot {A_2} \cdot {A_3} \ldots {A_n}} = {\bar A_1} + {\bar A_2} + {\bar A_3} + \ldots {\bar A_n}\)
\(\left( {ii} \right)\;\overline {{A_1} + {A_2} + {A_3} + \ldots + {A_n}} = {\bar A_1} \cdot {\bar A_2} \cdot {\bar A_3} \ldots {\bar A_n}\)
Application:
(a) F = ab + bc + ca
The inverse of the above function will be:
\(\bar F = \overline {\left( {ab + bc + ca} \right)} \)
Using Demorgan’s theorem, the above can be written as:
\(\bar F = \left( {\bar a + \bar b} \right) \cdot \left( {\bar b + \bar c} \right) \cdot \left( {\bar c + \bar a} \right)\)
\(\bar F = \bar a\bar b\bar c + \bar a\bar b + \bar b\bar c + \bar c\bar a\)
\(\bar F = \bar a\bar b \cdot \left( {\bar c + 1} \right) + \bar b\bar c + \bar c\bar a\)
Since 1 + any literal = 1 and 0 + any literal = literal, we get:
\(\bar F = \bar a\bar b + \bar b\bar c + \bar c\bar a\)
(b) F = ab + a̅ b̅ + c̅
\(\bar F = \overline {\left( {ab + \bar a\bar b + \bar c} \right)} \)
Again, Using the First form of DeMorgan’s Theorem, we get:
\(\bar F = \left( {\overline {ab} } \right) \cdot \overline {\left( {\bar a\bar b} \right)} \cdot \overline {\bar c} = \left( {\overline {ab} } \right)\;\overline {\left( {\bar a\bar b} \right)} .c\;\;\)
Using the 2^{nd} form of DeMorgan theorem, we get:
\(\bar F = \left( {\bar a + \bar b} \right)\left( {a + b} \right)\left( c \right)\)
\(\bar F = \left( {a\bar b + \bar ab} \right)\left( c \right) = \left( {a \oplus b} \right)c\)
Hence F̅ = (a ⊕ b)⋅ c
(c) F = a + bc
\(\bar F = \left( {\overline {a + bc} } \right) = \bar a \cdot \overline {bc} \)
\(\bar F = \left( {\bar a} \right) \cdot \left( {\bar b + \bar c} \right)\)
Hence, \(\bar F = \left( {\bar a} \right) \cdot \left( {\bar b + \bar c} \right)\)
(d) F = (a̅ + b̅ + c̅) (a + b̅ + c̅) (a̅ + b̅ + c)
\(\bar F = \overline {\left( {\bar a + \bar b + \bar c} \right)\left( {a + \bar b + \bar c} \right)\left( {\bar a + \bar b + c} \right)} \)
Applying DeMorgan’s Theorem two times, we get:
F̅ = a b c + a̅ bc + abc̅