Pseudo-Hadamard transform, most significant bit
The Pseudo-Hadamard Transform on $n$-bit integers $a$ and $b$ is defined as

$$\begin{align}
a^\prime &amp;= a + b &amp;\pmod {2^n}\\
b^\prime &amp;= a + 2b &amp;\pmod {2^n}
\end{align}$$

It is true that the most significant bit (MSB) of $b^\prime$ does not depend on the MSB of $b$. Whether or not this is a problem depends on exactly how it is used in the cipher. While I'm not aware of whether or not this makes any attacks possible on SAFER, Twofish, which also uses the PHT, considers this to be a non-issue. While the MSB is indeed lost, the reduction of security is insignificant, and any measures used to preserve the bit are not worth the performance impacts. From section 7.8 of the Twofish paper:


  The two outputs of the $g$ functions ($T_0$ and $T_1$) are then combined using a PHT so that both of them will affect both 32-bit Feistel XOR quantities. The half of the PHT involving the quantity $T_0 + 2T_1$ will lose the most significant bit of $T_1$ due to the multiply by two. This bit could be regained using some extra operations, but the software performance would be significantly decreased, with very little apparent cryptographic benefit. In general, the most significant byte of this PHT output will still have a non-zero output difference with probability $254/255$ over all 1-bit input differences.