Solution

The correct answer is (A), (B) Only

EXPLANATION:

• (A) {wi xj yk zl| i + k = j + l, where i, j, k, l ≥ 0}​ -
  • This language is context-free. It can be generated by a context-free grammar. For instance, we can create a grammar that generates strings where the number of 'w's plus the number of 'y's equals the number of 'x's plus the number of 'z's.
• (B) {wi xj yk zl| i = j and k = l, where i, j, k, l ≥ 0} -
  • This language is context-free. We can create a context-free grammar where 'w' and 'x' are generated together and 'y' and 'z' are generated together, ensuring i = j and k = l.
• (C) {wi xj yk zl| i = j = k and k ≠ l, where i, j, k, l ≥ 0} -
  • This language is not context-free. The constraint i = j = k imposes a condition that requires counting and comparing three separate counts, which cannot be achieved by a context-free grammar.
• (D) {wi xj yk zl| i = j = k + l, where i, j, k, l ≥ 0}​ -
  • This language is not context-free. The condition i = j = k + l requires a more complex relationship that cannot be captured by a context-free grammar.
• (E) {wi xj yk zl| i = j = l and k ≠ l, where i, j, k, l ≥ 0} -
  • This language is not context-free. Similar to the reason for (C), the conditions i = j = l and k ≠ l are too complex for a context-free grammar to generate.