= Solution
Use $T$ for the fixed maturity and $s\leq T$ for the observation time. The original PDF's discounting uses the <instantaneous forward rate> over $[s,T]$ and the <short rate> over $[0,s]$. Let $X(s,u)=F_{s,u}-\mu_{s,u}$ be the centered <Gaussian forward-rate field>, and define
$$
A(s,T)=\int_0^s\mu_{u,u}\,du+\int_s^T\mu_{s,u}\,du,\qquad
Y(s,T)=\int_0^T X(s\wedge u,u)\,du.
$$
Assume the <covariance> and mean have the regularity needed for these <mean-square integrals> and the maturity differentiation below. Splitting $Y$ at $u=s$ shows that the discounted <zero-coupon bond> price is
$$
Z_{s,T}=\exp[-A(s,T)-Y(s,T)].
$$
This <integrated Gaussian forward-rate process> has deterministic <variance>
$$
v(s,T)=\int_0^T\int_0^T c(s\wedge u\wedge w,u,w)\,du\,dw.
$$
Its <independent increments> hold relative to the entire specified <filtration>. Indeed, if $r\leq s$ and $z\leq r$, the <covariance> between $Y(s,T)-Y(r,T)$ and any earlier $X(z,w)$ is the integral of
$$
c((s\wedge u)\wedge z,u,w)-c((r\wedge u)\wedge z,u,w)=0.
$$
Since <uncorrelated jointly Gaussian variables are independent>, the increment is independent of every finite family of earlier observations and hence of their generated <sigma-algebra>. The increment has <normal distribution> with mean zero and variance $v(s,T)-v(r,T)$. Also $Y(0,T)=0$ because $c(0,u,w)=0$.
The <Gaussian moment-generating function> now gives
$$
\mathbb E[Z_{s,T}\mid\mathcal F_r]
=\exp[-A(s,T)-Y(r,T)+\tfrac12(v(s,T)-v(r,T))].
$$
Therefore the <Gaussian forward-rate covariance drift restriction> is exactly the deterministic condition
$$
\boxed{A(s,T)-A(0,T)=\frac12v(s,T).}
$$
It turns $Z_{s,T}/Z_{0,T}$ into a <Gaussian exponential martingale with deterministic variance>. The subparts establish the requested equivalent descriptions from this common calculation.
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