Brute force against computational security One-time pad is secure under brute force with respect to either of the definitions
\begin{align}
\Pr[M = m] &amp;= \Pr[M = m | C = c]\\
\Pr[C = c | M = m_0] &amp;= \Pr[C = c | M = m_1]
\end{align}

basically because the key and message spaces are the same, and $\oplus$ is closed under these spaces.

Given the argument above, I don't understand precisely how computationally secure schemes are compromised under brute force. In particular, how would any of the two equations above be violated given the scheme $$\mathsf{Enc}(m, k) = m \oplus G(k),$$ where $G$ is a pseudorandom generator.
For a (fixed-length) cipher to meet your first condition, it needs to be the case that it's no easier to guess the plaintext if you have the ciphertext than it is to guess the plaintext without the ciphertext. Now, suppose I send a random 1024-bit string XORed with $G(k)$ for some 128-bit $k$ and computationally secure $G$. The probability that my message is $0\ldots0$ is exactly $2^{-1024}$. I encrypt my message with a 128-bit key selected uniformly at random from the space of 128-bit keys to get $c$. Now suppose you intercept $c$. You know it's encrypted with some key $k$; you know there are $2^{128}$ such keys. That means that there are only $2^{128}$ possible decryptions -- at most 1 per key. This is why the keyspace of a perfectly secure scheme must be at least as big as the message space: knowing the ciphertext means that the number of possible messages is reduced to the number of possible keys, and if there are fewer keys than messages we've eliminated a lot of messages. That means that knowing "this ciphertext can decrypt to $m$" tells us that it's much more likely to be $m$ than we had suspected beforehand.

If $K=k$, then $M=c\oplus G(k)$. So, $$\Pr[M=c \oplus G(k)|C=c] \ge \Pr[K=k]=2^{-128}\ne 2^{-1024}$$ for any $k$. Suppose, for instance, that you try $k=1\ldots1$ and find that $c\oplus G(k)=0\ldots0$. You now know that $$\Pr[M=0\ldots0|C=c]\ge\Pr[K=1\ldots1]=2^{-128}\ne 2^{-1024}=\Pr[M=0\ldots0]$$ after a very quick computation (just one evaluation of $G$). That means that even with a computationally secure cipher, given $c$ an attacker can easily find a message $m$ such that $\Pr[M=m|C=c]\ge 2^{|k|}$ -- specifically, $\mathsf{Dec}(c,0...0)$. And that violates $\Pr[M=m|C=c]=\Pr[M=m]$ in a single evaluation of $\mathsf{Dec}$.