$L1= \{a^mb^nc^k\;|\;if\,(m=n)\,then\,(n!=k)\} \\ L2= \{a^ib^jc^k\;|\;if\,(i<j)\,then\,(k<j)\}\\ L3= \{a^ib^jc^k\;|\;(i<j)\,\leftrightarrow \,(k<j)\}$
Could someone please help to conclude the class of above languages? Below mentioned is the logic used to conclude that L1 and L2 are CFL but L3 is CSL but have noticed in most solutions L1 is treated as CSL. Could someone kindly point the flaw in below mentioned logic?
p implies q. If p then q. p -> q.
L1 = comlement(p) or q
= complement(m=n) or n!=k.
= m!=n or n!=k
L1 = { aabbccc, aabbbccc, aabbbcc.. } = NCFL
We can have union of two PDA's. One will accept the strings whenever m!=n and the other will accept the strings only when n!=k
L2 = comlement(p) or q
= complement( i< j ) or k<j.
= i>=j or k<j
L2 = { aabc, abbccc. } = NCFL
We can have union of two PDA's. One will accept the strings whenever i>=j and the other will accept the strings only when k<j
$L3 = p \leftrightarrow q\\ \;\; \overline{p}\;\overline{q}\;+\;pq$
= [ comp( i<j) and comp( k<j) ] or [ (i<j) and (k<j) ]
= [ (i>=j) and (k>=j) ] or [ ( i<j) and ( k<j) ]
= CSL