v_h)
 \\
 &   = \res{\uhk}{\z -  v_h} + \res{\uhk}{v_h} \quad \forall v_h \in \Vh,\notag\end{aligned}$$ where $\res{\uhk}{\cdot}$ is given by . Inserting $v_h:=\zhk$ in , we obtain $$\begin{aligned}
 \label{eq:JP_alg}
 \J(\u - \uhk) = \res{\uhk}{\z -  \zhk} + \res{\uhk}{\zhk}=: \errSk + \errAk,\end{aligned}$$ where the quantity $\errSk=\res{\uhk}{\z -  \zhk}$ represents the discretization error of the primal problem since it coincides with for $\uhk=\uh$. Further, the quantity $\errAk=\res{\uhk}{\zhk}$ represents the algebraic error of the primal problem since it vanishes for $\uhk=\uh$. Let us note that and hold without the Galerkin orthogonality .

In order to derive the analogue of including algebraic error, we take into account the error equivalence . The quantity $\ell_h(\z - \zhk)$ exhibits the analogue of the error $ \J(\u - \uhk)$ since both quantities are equal for $\uhk=\uh$ and $\zhk=\zh$. However, if $\uhk$ and $\zhk$ are arbitrary iterations then, in general, $$\begin{aligned}
  \label{eq:equi_alg}
  \J(\u - \uhk) \not= \ell_h(\z - \zhk).\end{aligned}$$ Nevertheless, we show in Section \[sec:algeb\] (cf. ) that if $\uhk$ and $\zhk$ are obtained by the bi-conjugate gradient (BiCG) method with vanishing initial approximations (in the exact arithmetic) then $$\begin{aligned}
  \label{eq:equi_algB}
  \J(\u - \uhk) =\ell_h(\z - \zhk)\qquad \mbox{ for } k=1,2,\dots.\end{aligned}$$

Now, using the consistency , identity $\z = 