There are 2^(2^n) different truth tables that can be constructed using the Boole

Question

There are 2^(2^n) different truth tables that can be constructed using the Boolean variables x1, . . . , xn, the operations , , ¬, and parentheses. What happens to this 2^(2^n) if we do not allow the use of the NOT operation “¬”? For n = 1 we only have one possible table, corresponding to the formula 1 = x1. For n = 2 we have four possibilities: 1 = x1, 2 = x2, 3 = x1 x2, or 4 = x1 x2. How many possibilities are there for n = 3? Give the number, list the corresponding formulæ, and explain how you derived this answer.

Explanation / Answer

Given n=3
so firstly we put in given formula
2^(2^n)===> 2^(2^3)===>2^8
which was almost 256 tables

Let us sort out without negation operator
1) Individually we can drae=w three tables
x1,x2,x3 (3)
x1 v x2 // x1 v x3 // x2 v x3 (3)
x1 ^ x3 // x1 ^ x3 // x2 ^ x3 (3)
(x1 v x2) v x3 // (x1 v x3) v x2 // (x2 v x3) v x1 (3)
similarlt and operator (3)
similarly by changing parantheses to right another (6)
OR and AND operator with all variables... (6)
  

so finally totally...>3+3+3+3+3+6+6==>27 tables

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