Produce the truth table of the combination of four NAND gates arranged as shown in Fig.Name the gate so formed.

Explanation

The truth table of combination of gates is shown below,
`|:{:(A,B,y'(=bar(A.B)),y_(1)(=bar(A.y')),y_(2)(=bar(B.y)),y=bar(y_(1).y_(2))),(0,0,1,1,1,0),(1,0,1,0,1,1),(0,1,1,1,0,1),(1,1,0,1,1,0):}:|-= |{:(A,B,y),(0,0,0),(1,0,1),(0,1,1),(1,1,0):}|`
It means a EXOR gate is produced.
Explaination
Using Boolean expression and De morgan's theorem [i.e., `bar(C.D)=barC+barD and bar(C+D)=barC.barD]`
`y^(')=bar(A.B)=barA+barB`
`y_(1)=bar(A.(barA+barB))=barA+bar((barA+barB))=barA+(bar(barA).bar(barB))=barA+(A.B)`
`y_(2)=bar(B.(barA+barB))=barB+bar((barA+barB))=barB+(bar(barA).bar(barB))=barB+(A.B)`
`y = bar(y_(1).y_(2)) =bar([bar(A)+(A.B)].[bar(B)+(A).B))] =bar([bar(A)+(A+B)]) =barbar(A).bar(A+B)+barbar(B).bar(A+B)`
`=A.(barA+barB)+B.(barA+barB)=A.barA+A.barB+B.barA+B.barB=A.barB+B.barA ( :'A.barA=B.barB=0)`
which is for EXOR gate.
Thus, arrangement works as EXOR gate.