Simplifying output for a XOR gate using Boolean Algebra

[Potatochip911]

Homework Statement

I'm trying to show that the output of this XOR circuit is $F=A'B+AB'$,

Homework Equations

$(A+B)'=A'\cdot B'$
$(A\cdot B)'=A'+B'$

The Attempt at a Solution

From the gates the output is $[(A\cdot B)+(A+B)']'$, using De Morgan's laws this becomes $[(A\cdot B)+(A+B)']'=(A\cdot B)'\cdot (A+B)=(A\cdot B)'\cdot (A+B)=(A'+B')\cdot (A+B)=0$? I can't seem to figure out what I'm doing wrong.

 

[ehild]

Homework Statement

I'm trying to show that the output of this XOR circuit is $F=A'B+AB'$,

Homework Equations

$(A+B)'=A'\cdot B'$
$(A\cdot B)'=A'+B'$

The Attempt at a Solution

From the gates the output is $[(A\cdot B)+(A+B)']'$, using De Morgan's laws this becomes $[(A\cdot B)+(A+B)']'=(A\cdot B)'\cdot (A+B)=(A\cdot B)'\cdot (A+B)=(A'+B')\cdot (A+B)=0$? I can't seem to figure out what I'm doing wrong.

You did it right, but simplify the last expression applying the distributive law.

 

[Potatochip911]

You did it right, but simplify the last expression applying the distributive law.

yikes, can't believe I missed that one!

 

[ehild]

yikes, can't believe I missed that one!

Never give up hope :)