Modify the PDA in 4.2.1.4 to accept {0n 1n : n ≥ 1} by ES.
Modify the PDA in 4.2.1.4 to accept {0n 1n : n ≥ 3} by ES.
Build a PDA over an input alphabet ∑ that accepts strings by ES, and (provably) accepts {xxR : x ∑*}.
Build a PDA for the language {0n15n : n ≥ 0}. You are free to choose the mode of acceptance.
Repeat the above task, but now do it in this way: First get a CFG for the language, and then construct a PDA for the language that accepts by empty stack (cf. 4.3.0.5).
Give a CFG G over ∑ = {0, 1} such that L(G) = {xxR : x ∑*}.
Give a CFG G over ∑ = {(, )} such that L(G) is the full set of balanced brackets—that is, not only those of the form ((())) but also those like ((())) (()).
By induction on regular expression length prove: “For every α, there is a CFG, G, such that L(G) = L(α)”.
Prove that CFL are closed under union, that is, if L and L′ are CFL, then so is L ∪ L′.
Prove that CFL are closed under concatenation, that is, if L and L′ are CFL, then so is LL′.
Prove that CFL are closed under reversal, that is, if L is a CFL, then so is LR.
Prove that CFL are not closed under complement. That is, if L is a CFL over ∑, then it is not necessarily the case that , that is, ∑* − L is a CFL. ...
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