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The given gates are XNOR gate.

At the first gate we can see that we have P **XNOR **0, which gives 1 if p=0 and 0 if p=1 so we can say that our output from the first gate is **P' **(P complement).

Now this output is fed as a input to the next gate , so we have P ⊙ P' at the next gate. Now P ⊙ P' means the combination will always be either 0 ⊙ 1 or 1 ⊙ 0, so this gate will always give output as 0.

So, i/p to the 3rd gate is 0 ===> which is seen like original problem with 22-2 = __ 20__ XNOR gates

So, after 2 more gates o/p again 0.

Finally, after 2 gates, after 4 gates, after 6 gates....... o/p is 0.

there are total 22 gates which are even ===> o/p at the end should be 0

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