The following expression was to be realized using 2-input AND and OR gates.…

Key idea: replacing every 2-input AND by a 2-input NAND changes chained ANDs. For a chain (x.y).z implemented with two 2-input NANDs we get NAND(NAND(x,y),z) = NOT((NOT(xy)) AND z) = xy + z'.

• Transform each grouped term using NAND replacements:

• (a.b).c becomes ab + c' (since NAND(NAND(a,b),c) = ab + c').

• (a'.c).d becomes a'c + d'

• (b.c).d becomes bc + d'

• a.d (single 2-input AND) becomes a' + d' (since NAND(a,d) = a' + d').

Combine all transformed terms (OR them):

F = (ab + c') + (a'c + d') + (bc + d') + (a' + d')

Simplify by removing redundancies and using absorption:

• Collect terms: F = a' + c' + d' + ab + a'c + bc

• a'c is redundant because a' is already present, so drop a'c.

• Use c' + bc = c' + b (absorption), so ab + (c' + bc) simplifies to ab + (c' + b) = b + c' (since b + ab = b).

• Therefore F simplifies to a' + b + c' + d'

Final answer: a' + b + c' + d'.