SOLUTION: give a proof of ⊢ A V B ≡ A V ¬B ≡ A

You must put parentheses and brackets in logic statements to determine
which statements (indicated by letters A,B) go with which connectives v,≡,¬.
Otherwise the compound statement could mean different things.  Parentheses 
are to indicate which order to do things in.  Without them there is no way
to tell, because there are no rule of hierarchy of connectives as there are 
in of hierarchy with addition, subtraction multiplication and division of
ordinare mathematics.  So I'm guessing that you mean this:

[(A V B) ≡ (A V ¬B)] ≡ A

Put TTFF under the A's and TFTF under the B's

[(A V B) ≡ (A V ¬B)] ≡ A
  T   T     T    T     T
  T   F     T    F     T
  F   T     F    T     F
  F   F     F    F     F

This method of truth tables starts with columns of truth values 
under only the letters. Then we get letters under the connectives
erasing the columns under the letters each time.  We eventually
wind up with just one column of truth values under the connective 
that is not enclosed in parenthese or brackets.

Under the ¬ before the B list the opposite of what's under
the B next to it and then erase what's under the B:

[(A V B) ≡ (A V ¬B)] ≡ A
  T   T     T   F      T
  T   F     T   T      T
  F   T     F   F      F
  F   F     F   T      F

Under the first v place F only where there are F's on both
sides of the v, and T's elsewhere. Then erase what's under 
A and B:

[(A V B) ≡ (A V ¬B)] ≡ A
    T       T   F      T
    T       T   T      T
    T       F   F      F
    F       F   T      F

Under the second v

Under the second v place F only where there are F's on both
sides of the v, and T's elsewhere. Then erase what's under 
A and the ¬:

[(A V B) ≡ (A V ¬B)] ≡ A
    T         T        T
    T         T        T
    T         F        F
    F         T        F

Under the ≡ place T if what's on both sides of it are the same,
and F if they are not. Then erase what's under the two v's.

[(A V B) ≡ (A V ¬B)] ≡ A
         T             T
         T             T
         F             F
         F             F

Now under the second ≡ place T if what's on both sides of it are the 
same, and F if they are not. Then erase what's under the first ≡ and
the A:

[(A V B) ≡ (A V ¬B)] ≡ A
                     T
                     T
                     T
                     T

We wound up with all T's and no F's under the connective ≡ that is not 
inside any parentheses or brackets.

That proves the compound statement [(A V B) ≡ (A V ¬B)] ≡ A is always true,
and we say "It is a tautology".

Edwin