Consider the language \( L = \\{ 0^n \\mid n \\text{ is a prime number} \\} \\) and we want to use the pumping lemma to prove that \( L \\) is not a regular language. Assume \( L \\) is a regular language and let \( z = 0^p \\) where \( p \\) is a prime number greater than the pumping length \( N \\). According to the pumping lemma, there exists a string \( z = uvw \\) that satisfies certain conditions. Which of the following options correctly describes the key step in this proof?

Options:
A. Assume \( u = 0^{p-k-j} \\) and \( v = 0^k \\) (\( k \\geq 1 \\)), and \( w = 0^j \\). Then, when \( i = N + p + 1 \\), the string \( uv^{N+p+1}w = 0^{(N+p)(1+k)} \\). Since \( k \\geq 1 \\) and \( N + p + (N + p)k \\) is not a prime number, \( uv^{N+p+1}w \\notin L \\). This contradicts the pumping lemma, thus proving that \( L \\) is not a regular language.
B. Assume \( v = 1^k \\) (\( k \\geq 1 \\)). Then, for all \( i \\geq 0 \\), the string \( uv^iw \\) belongs to \( L \\), which is consistent with the definition of \( L \\), and therefore cannot prove that \( L \\) is not a regular language.
C. Assume \( u = 0^{p-k-j} \) and \( v = 0^k \) (\( k \geq 1 \)), and \( w = 0^j \). Then, when \( i = 2 \), the string \( uv^2w = 0^{p+2k} \). Since \( k \geq 1 \) and \( p + 2k > p \), \( uv^2w \notin L \). This contradicts the pumping lemma, thus proving that \( L \) is not a regular language.
D. Assume \( u = 0^{p-k-j} \) and \( v = 0^k \) (\( k \geq 1 \)), and \( w = 0^j \). Then, when \( i = N + 1 \), the string \( uv^{N+1}w = 0^{p + Nk} \). Since \( k \geq 1 \) and \( p + Nk \) is not a prime number, \( uv^{N+1}w \notin L \). This contradicts the pumping lemma, thus proving that \( L \) is not a regular language.
E. Assume \( u = 0^{p-k-j} \) and \( v = 0^k \) (\( k \geq 1 \)), and \( w = 0^j \). Then, when \( i = p \), the string \( uv^pw = 0^{p+k(p-1)} \). Since \( k \geq 1 \) and \( p + k(p-1) \) is not a prime number, \( uv^pw \notin L \). This contradicts the pumping lemma, thus proving that \( L \) is not a regular language.
F. Assume \( u = 0^{p-k-j} \) and \( v = 0^k \) (\( k \geq 1 \)), and \( w = 0^j \). Then, when \( i = p + 1 \), the string \( uv^{p+1}w = 0^{p + (p+1)k} \). Since \( k \geq 1 \) and \( p + (p+1)k \) is not a prime number, \( uv^{p+1}w \notin L \). This contradicts the pumping lemma, thus proving that \( L \) is not a regular language.
G. Assume \( u = 0^{p-k-j} \\) and \( v = 0^k \\) (\( k \\geq 1 \\)), and \( w = 0^j \\). Then, when \( i = 2 \\), the string \( uv^2w = 0^{p+k} \\). Since \( k \\geq 1 \\) and \( p + k > p \\), \( uv^2w \\notin L \\). This contradicts the pumping lemma, thus proving that \( L \\) is not a regular language.
H. Assume \( v = 0^k \\) (\( k \\geq 1 \\)). Then, for all \( i \\geq 0 \\), the string \( uv^iw \\) belongs to \( L \\), which is consistent with the definition of \( L \\), and therefore cannot prove that \( L \\) is not a regular language.
I. Assume \( u = 0^{p-k-j} \) and \( v = 0^k \) (\( k \geq 1 \)), and \( w = 0^j \). Then, when \( i = p - 1 \), the string \( uv^{p-1}w = 0^{p + k(p-2)} \). Since \( k \geq 1 \) and \( p + k(p-2) \) is not a prime number, \( uv^{p-1}w \notin L \). This contradicts the pumping lemma, thus proving that \( L \) is not a regular language.
J. Assume \( u = 0^{p-k-j} \) and \( v = 0^k \) (\( k \geq 1 \)), and \( w = 0^j \). Then, when \( i = N + 1 \), the string \( uv^{N+1}w = 0^{p + N + 1} \). Since \( k \geq 1 \) and \( p + N + 1 \) is not a prime number, \( uv^{N+1}w \notin L \). This contradicts the pumping lemma, thus proving that \( L \) is not a regular language.
Answer: A