Question

x <-> y1 NAND y2 ;

in this case how to write the truth table value for this entire expression?

Answer 1

What is the problem? The formula (y1 NAND y2) is syntactic sugar for !(y1 & y2), so that we have to compute the truth table for x <-> !(y1 & y2) which is the following:

x	y1	y2	x <-> !(y1 & y2)
0	0	0	0
0	0	1	0
0	1	0	0
0	1	1	1
1	0	0	1
1	0	1	1
1	1	0	1
1	1	1	0

Comments

how to calculate <-> ?

I am aware that y1 nand y2 is -(y1 and y2)

---

You simply compare the values of !(y1 & y2) with the values of x. If they are the same, the resulting value is 1 (because they are equivalent), otherwise the resulting value is 0 (because both parts are not equivalent). Or you can simply rewrite it as:
a <-> b = a&b | !a&!b = (a->b) & (b->a) = (!a|b) & (!b|a)
All four formulas describe the same Boolean function, which calculates the equivalence of two values a, b.