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This formal language is not a context-free language. For pumping length $p$, use $a^p{b}^{p^2}{c}^p$. A pumpable window of length at most $p$ cannot meet both outer blocks. If pumping adds $\alpha ,\beta ,\gamma$ copies of $a,b,c$, then $\alpha \gamma =0$ and $1\le \alpha +\beta +\gamma \le p$. Preserving the product constraint would require $\beta =p(\alpha +\gamma )$, impossible both when $\alpha +\gamma =0$ and when it is positive. This contradicts the pumping lemma for context-free languages.

No: regular languages are not closed under countably infinite intersection. Let $\Sigma =\{0,1\}$ and putThe first term in $L_i$ has regular expression $(0\cup 1{)}^i(0\cup 1{)}^{\ast }$, where the fixed power $i$ means $i$ copies. The second term is a finite language, a finite union of single words. Their union is therefore a regular language.

If $w\in K$, then for each $i$ either $|w|\ge i$ or $w$ belongs to the second term, so $w\in L_i$. If $w\notin K$, choose $i>|w|$. It belongs to neither term of $L_i$. ConsequentlyThe formal language $K$ is not a regular language. Indeed, for a proposed pumping length $p\ge 1$, the pumping lemma for regular languages applied to $0^p{1}^p$ must pump a nonempty block consisting only of zeros. Repeating that block increases the number of zeros without increasing the number of ones, leaving $K$. This contradiction proves the counterexample. Finite intersection closure does not extend to this infinite intersection of regular languages.

Every generated word satisfieswhere ${\#}_a,{\#}_b$ count occurrences of the indicated terminal symbols. To see this, follow the unique chain of $S$ occurrences in a parse tree. Each use of $S\to UST$ introduces one $a$ from $U$, at least one $b$ from its rightmost $T$, and possibly one further $b$ from $U\to aT$. Each use of $S\to ST$ adds one $b$ and no $a$. The terminating $S\to \epsilon$ contributes neither. Hence no generated word has more $a$ symbols than $b$ symbols.

Also, for every $n\ge 0$, the word $a^n{b}^n$ is generated: apply $S\to UST$ exactly $n$ times, choose $U\to a$ each time, then use $S\to \epsilon$ and all remaining $T\to b$.

Suppose $L(G)$ were a regular language, and let $p\ge 1$ be a pumping length from the pumping lemma for regular languages. Apply it to $w=a^p{b}^p$. Every allowed decomposition $w=xyz$ with $|xy|\le p$ and $|y|>0$ has $y=a^r$ for some $r\ge 1$. Pumping twice giveswhich violates the proved counting inequality and hence is not generated. This contradicts the pumping lemma for regular languages. Therefore $L(G)$ is not a regular language. Ambiguity was not used: the obstruction is the unbounded constraint on symbol counts.

The formal language is not a context-free language. Suppose otherwise, and let $p\ge 1$ be a pumping length from the pumping lemma for context-free languages. ChooseThis word belongs to the formal language, since its middle exponent is the product of the two outer exponents. The pumping lemma for context-free languages would givewith $u{v}^{i}x{y}^{i}z$ in the formal language for every integer $i\ge 0$.

Let $\alpha ,\beta ,\gamma$ be the respective numbers of $a,b,c$ symbols in the two pumped pieces $v,y$ together. The window $vxy$ cannot contain both an $a$ and a $c$: it would have to contain the entire middle block of length $p^2$, plus at least those two outer symbols, exceeding $p$. HencePump once extra, taking $i=2$. The resulting symbol counts are $p+\alpha$, $p^2+\beta$, and $p+\gamma$. A necessary condition for membership, regardless of whether the block order survives, iswhere $\alpha \gamma =0$ was used.

If $\alpha +\gamma =0$, then $\beta >0$, contradicting that equality. If $\alpha +\gamma \ge 1$, thenagain contradicting it. Thus every permissible decomposition fails at $i=2$. This contradicts the pumping lemma for context-free languages and proves the claim. The obstruction is the multiplicative block-count language: a bounded local change cannot preserve the product constraint.

Let $K$ be the diagonal halting set, which is undecidable. The pumping argument below actually holds for every subset $K\subseteq \mathbb{N}$; nonregularity needs the stated convention, or at least a set of exponents whose unary language is nonregular.

The pumping lemma for regular languages holds with pumping length one. For an accepted nonempty word, pump its first symbol. A word containing only $1$s remains in $1^{\ast }$ even if that symbol is deleted. If there is a zero and a nonempty prefix before its last zero, pumping the first symbol leaves the last zero and its following $n\in K$ ones unchanged. If the word is $01^n$, pumping its initial zero produces $0^i{1}^n$; this is accepted for $i\ge 1$, and for $i=0$ it belongs to $1^{\ast }$. In every case the pumped substring is nonempty and lies in the first position.

If $L$ were a regular language, closure under intersection would makeregular. A finite-state automaton for this intersection would decide membership of $n$ in the diagonal halting set by reading $01^n$, a contradiction. Thus $L$ satisfies the pumping lemma but is not regular, under the usual meaning of $\mathbb{K}$. If $K$ were arbitrary, the printed conclusion would be false, for example for $K=\mathbb{N}$.

The pumping lemma for context-free languages states that every context-free language $L$ has a pumping length $p$ such that every $w\in L$ with $|w|\ge p$ can be writtenwith $|vxy|\le p$, $|vy|>0$, and $u{v}^{i}x{y}^{i}z\in L$ for every integer $i\ge 0$.

This language is not context-free. Let $p$ be a proposed pumping length and choose $n$ with $3^n\ge p$. In any pumping decomposition of the unary word $a^{3^n}$, pumping once more increases its length by an integer $d$ with $1\le d\le p$. Butso the pumped length is not a power of three, contrary to the pumping lemma for context-free languages. This is also an instance of the fact that a unary context-free language is regular, whereas the powers of three are not ultimately periodic.

LetIntersect it with the regular languageEvery word of $R$ contains exactly two copies of $b$. If it is a proper power, its exponent must therefore be two and its root must contain one $b$. HenceThis language is not regular: if $p$ were a pumping length, pumping any nonempty substring among the first $p$ copies of $a$ in $a^{p}b{a}^{p}b$ would change only its first $a$-block. Since regular languages are closed under intersection,This is the proper-power language is not regular.

The language $L=\{a^{m^2}{b}^{m^2}:m\ge 0\}$ is not context-free. Suppose it had pumping length $p$ in the pumping lemma for context-free languages, and apply the lemma to $w=a^{p^2}{b}^{p^2}$. Write $w=uvxyz$ with $|vxy|\le p$, $|vy|>0$, and $u{v}^{i}x{y}^{i}z\in L$ for every $i\ge 0$.

Let $\alpha$ and $\beta$ be the respective numbers of $a$s and $b$s in $vy$. If $\alpha \ne \beta$, pumping with $i=0$ or $i=2$ destroys equality of the two block lengths. Hence a valid decomposition would require $\alpha =\beta =d>0$. Pumping with $i=2$ would then produce $a^{p^2+d}{b}^{p^2+d}$, where $1\le d\le p$. Butso $p^2+d$ is not a square number. This contradiction proves the square-count language is not context-free.

The pumping lemma for context-free languages rules out a formal language when two pumpable pieces cannot preserve all three linked block counts. For example, let $L=\{a^n{b}^n{c}^n:n\ge 0\}$. If $L$ were a context-free language with pumping length $p$, the word $a^p{b}^p{c}^p$ would have a decomposition $uvxyz$ with $|vxy|\le p$ and $|vy|>0$.

The window $vxy$ cannot contain both an $a$ and a $c$, because the intervening $b$ block alone has length $p$. Pumping down to $uxz$ therefore leaves at least one outer block at length $p$, while deleting at least one symbol from another block. The remaining block counts cannot all be equal, so $uxz\notin L$, contrary to the pumping lemma for context-free languages. This argument uses a necessary condition for context-free languages; satisfying a pumping condition would not itself prove that a formal language is a context-free language.