Proof Systems, implication of complete and soundness of itself negated

Question

Let $A(S,P,f,g)$ and $B=(S,P,h,i)$ $S$ is the sound set of statements for $A$ and $B$ $P$ is the sound set of proofs for $A$ and $B$ Define $h$ and $i$ as:$$\begin{align}h(s)=1\iff&\,p(s)=0\\i(s,p)=1\iff&\,g(s,p)=0\end{align}$$Prove or disprove:

1. If $A$ is sound, then $B$ is complete
2. If $A$ is complete, then $B$ is sound

Looking at the two Proof System, we could say that $B$ is $A$ negated (the truth function and the verification function of $B$ give the exact opposite of the one in $A$).

As shown in the picture there are two statement to be proven i tried to solve them and my answer appeared to be true for both. But one of my friend said that I can prove that the second is false with a counter example, he thought:

Since A is complete we have: $f(s_1) = 1$ then $g(s_1,p_1) = 1$

but since nothing is said about the soundness of A in this part he said also that:

$f(s_2) = 0$ then $g(s_2, p_2) = 1$

starting from this he proved that not necessarily if $A$ is complete, then $B$ is sound.