Correct answer is L2 and L4
EXPLANATION :
L1: 2 comparisons so not CFL
L2: 1 comparison so CFL
if you look through it properly you will see that the union (i<=j) U (j<=i) means i and j can be anything(since i can be less than or equal to j OR j can be less than or equal to i). So there is just one comparison.
L3: more than 1 comparison so not CFL |b| = |c| , |a| != |b| , |c| != |d|
L4: 1 comparison so CFL
p --> q which is equivalent to (not p or q )
where
p : |a| = |b|
q : |c| is even
if p is false , directly accept the string
if p is true, check for even no. of c which can be done by DFA
