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Every regular language is also a context-free language.

The language $L = \{ a^n b^n c^n| n \geq 0 \} \,$ L = \{ a^n b^n c^n| n \geq 0 \} \, is a context free language

A context free grammar is a 4-tuple $G=( V, \Sigma, R, S )$ G=( V, \Sigma, R, S ) where

$\Sigma$ \Sigma is a finite set of terminals

$\Sigma$ \Sigma is a infinite set of terminals

$S\in V$ S\in V is not the start rule

V is an infinite set of variables

Nondeterministic pushdown automata recognize certain languages that no deterministic pushdown automata can recognize.

Deterministic and nondeterministic pushdown automata are equivalent.

If A is a context-free language, then there is a number p where, if $s\in A$ s\in A of length at least p, then s may be divided into five pieces s = uvxyz satisfying the following condition

for $i\geq 0,$ i\geq 0, $\,\, v^ixy^i \in A$ \,\, v^ixy^i \in A

for $i\geq 0,$ i\geq 0, $\,\, uv^ixy^iz \in A$ \,\, uv^ixy^iz \in A

for $i\geq 0,$ i\geq 0, $\,\, uvx^iyz \in A$ \,\, uvx^iyz \in A

for $i\geq 0,$ i\geq 0, $\,\, uv^ix^iy^iz \in A$ \,\, uv^ix^iy^iz \in A

The regular expression $(0 1^* 0 \cup 1 )^* 0$ (0 1^* 0 \cup 1 )^* 0 generates the language consisting of all strings over $\{ 0, 1 \}$ \{ 0, 1 \} having and odd number of 0s