Step 1  Simplify the given expression using appropriate theorems/rules.
           \(\bar { X } .Y+Y.(\bar { Z } (\bar { Z } +Y))=\bar { X } .Y+Y.(\bar { Z } .\bar { Z } +\bar { Z } .Y)\quad \quad \quad \quad [By\quad using\quad Distributive\quad law]\\ =\bar { X } .Y+Y.(\bar { Z } +\bar { Z } .Y)\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad [\because \quad \bar { Z } .\bar { Z } =\bar { Z } ]\\ =\bar { X } .Y+Y.\bar { Z } .(1+Y)\\ =\bar { X } .Y+Y.\bar { Z } .1\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad [\because \quad 1+Y=1]\\ =\bar { X } .Y+Y.\bar { Z } \\ \\ \)
Step 2 To convert into POS form applying the boolean algebra rule
         \(\bar { X } .Y+Y.\bar { Y } =(\bar { X } .Y+Y).(\bar { X } .Y+\bar { Z } )\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad [\because \quad X+Y.Z=(X+Y).(X+Z)]\\ =(Y+\bar { X } .Y).(\bar { Z } +\bar { X } .Y)\quad \quad [\because X+Y=Y+X]\\ =(Y+\bar { X } ).(Y+Y).(\bar { Z } +\bar { X } ).(\bar { Z } +Y)\\ =(\bar { X } +Y).(Y).(\bar { X } +\bar { Z } ).(Y+\bar { Z } )\quad \quad [\because \quad Y+Y=Y)\)
Step 3 Wherever a variable is missing, ad that term with (missing variable. complement of missing variable)
        \(=(\bar { X } +Y+Z.\bar { Z } ).(X.\bar { X } +Y+Z.\bar { Z } ).(\bar { X } +Y.\bar { Y } +\bar { Z } ).(X.\bar { X } +Y+\bar { Z } )\\ =(\bar { X } +Y+Z).(\bar { X } +Y+\bar { Z } ).(X.\bar { X } +Y+Z).(X.\bar { X } +Y+\bar { Z } ).(\bar { X } +Y+\bar { Z } ).(\bar { X } +\bar { Y } +\bar { Z } ).((X+Y+\bar { Z } ).(\bar { X } +Y+\bar { Z } )\\ =(\bar { X } +Y+Z).(\bar { X } +Y+\bar { Z } ).(X+Y+Z).(\bar { X } +Y+Z).(X+Y+\bar { Z } ).(\bar { X } +Y+\bar { Z } ).(\bar { X } +Y+\bar { Z } ).(\bar { X } +\bar { Y } +\bar { Z } ).((X+Y+\bar { Z } ).(\bar { X } +Y+\bar { Z } )\\ \) 
Step 4 Remove all duplicate terms, i.e.,
       \(=(\bar { X } +Y+Z).(\bar { X } +Y+\bar { Z } ).(X+Y+Z).(X+Y+\bar { Z } ).(\bar { X } +\bar { Y } +\bar { Z } )\\ \) 
This is the desired canonical Product of Sums form.

Sum of Products(SOP) expression contain sum of all minterms for output as 1.
    \(=\bar { A } .\bar { B } .\bar { C } +\bar { A } .\bar { B } .C+\bar { A } .B.C+A.\bar { B } .C+A.B.\bar { C } \) 
    \(=\sum (0,1,3,5,6)\)
Product of Sum(POS) expression contains product of all maxterms for output as 0.
      \(=(A+\bar { B } +C).(\bar { A } +B+C).(\bar { A } +\bar { B } +\bar { C } )\quad \)
      \(=\prod (2,4,7)\)

F= \(A.\bar { B } .C+\bar { A } .B.C+\bar { A } .B.\bar { C } \) 
i.e., (101)+(011)+(010)=m₅+m₃+m₂=\(\sum (2,3,5)\) 
POS expression will be : 
          \(=\prod (0,1,4,6,7)\) 
          = M₀.M₁.M4.M₆.M₇
          =(000).(001).(100).(110).(111)
          =\((A+B+C).(A+B+\bar { C } ).(\bar { A } +B+C).(\bar { A } +\bar { B } +C).(\bar { A } +\bar { B } +\bar { C } )\)

\(\bar { X } .Y+X.\bar { Y } =(\bar { X } +\bar { Y } ).(X+Y)\) 
Taking R.H.S \((\bar { X } +\bar { Y } ).(X+Y)\) 
                         \(=\bar { X } .X+\bar { X } .Y+\bar { Y } .X+\bar { Y } .Y\) 
                         \(=0+\bar { X } .Y+\bar { Y } .X+0\quad \quad \quad \quad \quad \quad [\therefore \quad X.\bar { X } =0]\) 
                         \(=\bar { X } .Y+\bar { Y } .X\)                                \([\bar { X } .Y=\bar { Y } .X]\)  
                          = L.H.S        Hence proved.

\(A+C=A+\overset { - }{ A } .C+B.C\)

RHS=\(A+\overset { - }{ A } .C+B.C\)
\(A+C=A+\overset { - }{ A } .C+B.C\\ \quad \quad \quad \quad =(A.\overset { - }{ A } ).(A+C)+B.C\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad [By\quad using\quad distributive\quad law]\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad [By\quad using\quad A+\overset { - }{ A } =1]\\ \quad \quad \quad \quad =1.(A+C)+B.C\\ \quad \quad \quad \quad =A+C+B.C=A+C.(1+B)\\ \quad \quad \quad \quad =A+C.(1)=A+C\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad [\therefore 1+B=1]\\ \quad \quad \quad \quad =LHS\)