Prove that F(a, b, c) = π(2, 3, 4, 7) = Σ(0, 1, 5, 6).

F(a, b, c) = π(2, 3, 4, 7)
Reducing using K-Map:

From Pair (3,2):
Rows representing the Pair: a
Columns representing the Pair: (b'+c').(b'+c) = b'
Term Obtained = a+b'

From Pair (3,7):
Rows representing the Pair: 1 (Both a and a' cancel each other)
Columns representing the Pair: b'+c'
Term Obtained = b'+c'

From (4):
Row representing (4): a'
Columns representing (4): b+c Term Obtained = a'+b+c

Reduced Expression = (a + b').(b' + c').(a' + b + c)

F(a, b, c) = Σ(0, 1, 5, 6)
Reducing using K-Map:

From Pair (0,1):
Rows representing the Pair: a'
Columns representing the Pair: b'c' + b'c = b'
Term Obtained = a'b'

From Pair (1,5):
Rows representing the Pair: 1 (Both a and a' cancel each other)
Columns representing the Pair: b'c
Term Obtained = b'c

From (6):
Row representing (6): a
Columns representing (6): bc' Term Obtained = abc'

Reduced Expression = a'b' + b'c + abc'

To prove:
(a + b').(b' + c').(a' + b + c) = a'b' + b'c + abc'

LHS = (a' + b + c).(a + b').(b' + c')
    = (aa' + a'b' + ab + bb' + ac + b'c).(b' + c')
    = (a'b' + ab + ac + b'c).(b' + c')
    = a'b'b' + a'b'c' + abb' + abc' + ab'c + acc' + b'b'c + b'cc'
    = a'b' + a'b'c' + 0 + abc' + ab'c + 0 + b'c + 0
    = a'b' + a'b'c' + abc' + ab'c + b'c
    = a'b'(1 + c') + abc' + b'c(a + 1)
    = a'b' + abc' + b'c
    = RHS

Hence Proved.