Dear Amal,

Salam,

In the attached file, the formulas precisely prove why Pr[ no E3] >=1/2. E3 is an event based on which the adversary A guesses the bit b'. If the attack applied by A is succeeded, its advantage is not negligible, means that 

|Pr[b=b']-1/2| >= epsilon.       (1)

Since the guess of b' is made according to the event E3 (described in the previous pages of the attached file), we can use of Total Probability Theorem to write

Pr[b=b'] = Pr[b=b'|E3]Pr[E3] + Pr[b=b'|no E3]Pr[no E3]      (2)

Since the value of the term "Pr[b=b'|no E3]" is between 0 and 1, then

Pr[b=b'|no E3]Pr[no E3] == 1/2Pr[E3]         (8)

if we replace Pr[E3] by "1-Pr[no E3]", we have

Pr[b=b'] >= 1/2 - 1/2Pr[no E3]      (9)

The above equation can be written as 

Pr[b=b'] - 1/2 >= -1/2Pr[no E3]       (10)

From the equations (6) and (10) we have 

|Pr[b=b'] - 1/2| =< 1/2Pr[no E3]         (11)

By combining the equations (1) and (11) we have 

epsilon =< |Pr[b=b'] - 1/2| =< 1/2Pr[no E3]                (12)

As our goal was to find a lower bound for Pr[no E3], we obtain from the equation (12) that 

epsilon =< 1/2Pr[no E3] 

and then 

2epsilon =< Pr[no E3]

that is the end of our proof to find a lower bound for Pr[no E3].