[proofplan]
The proof is a direct extraction of the necessary optimality equations for a Lasso minimizer with the prescribed signed support. On the active coordinates, the differentiable quadratic part balances the $\ell^1$ subgradient $s$, and invertibility of $\hat{\Sigma}_{SS}$ gives an explicit formula for $\hat{\beta}_S$. On the inactive coordinates, the one-sided directional optimality condition gives the dual bound, and substituting the active formula produces the stated residual-noise vector. The probability bound follows from event inclusion, and the final obstruction follows from the [reverse triangle inequality](/theorems/2300).
[/proofplan]
[step:Derive the coordinatewise optimality conditions for the assumed minimizer]
We use the componentwise sign convention $\operatorname{sign}(0)=0$. For each $j\in\{1,\dots,p\}$, let $X_j\in\mathbb R^n$ denote the $j$th column of $X$. Define the Lasso objective $Q_\lambda: \mathbb{R}^p \to \mathbb{R}$ by
\begin{align*}
Q_\lambda(b)
:=
\frac{1}{2n}\|Y-Xb\|_2^2+\lambda\|b\|_1.
\end{align*}
If $S=\varnothing$, all active-coordinate equations below are interpreted as empty vector identities, and the inactive-coordinate argument remains unchanged. Fix a realization of $\varepsilon$ for which there exists a minimizer $\hat{\beta} \in \operatorname*{arg\,min}_{b \in \mathbb{R}^p} Q_\lambda(b)$ satisfying $\operatorname{supp}(\hat{\beta})=S$ and $\operatorname{sign}(\hat{\beta}_S)=s$. Let $r := Y-X\hat{\beta} \in \mathbb{R}^n$ denote the residual vector. For each $j \in S$, let $e_j \in \mathbb{R}^p$ denote the $j$th standard basis vector. The function $t \mapsto Q_\lambda(\hat{\beta}+t e_j)$ is differentiable at $t=0$, because $\hat{\beta}_j \ne 0$ and the absolute value term is differentiable there. Since $t=0$ is a minimum, its derivative vanishes:
\begin{align*}
-\frac{1}{n}X_j^\top r+\lambda \operatorname{sign}(\hat{\beta}_j)=0.
\end{align*}
Thus, using $\operatorname{sign}(\hat{\beta}_S)=s$,
\begin{align*}
\frac{X_S^\top r}{n}=\lambda s.
\end{align*}
For each $j \in S^c$, we have $\hat{\beta}_j=0$. Since $t=0$ minimizes $t \mapsto Q_\lambda(\hat{\beta}+t e_j)$, for every $t \in \mathbb{R}$,
\begin{align*}
0
\le
Q_\lambda(\hat{\beta}+t e_j)-Q_\lambda(\hat{\beta})
=
-\frac{t}{n}X_j^\top r
+
\frac{t^2}{2n}|X_j|^2
+
\lambda |t|.
\end{align*}
Dividing by $t>0$ and letting $t \downarrow 0$ gives $-\frac{1}{n}X_j^\top r+\lambda \ge 0$, since the right-hand side after division is continuous in $t$ at $0$. Dividing by $t<0$ and letting $t \uparrow 0$ gives $-\frac{1}{n}X_j^\top r-\lambda \le 0$, again by continuity at $0$. Therefore
\begin{align*}
\left|\frac{X_j^\top r}{n\lambda}\right|\le 1.
\end{align*}
Equivalently,
\begin{align*}
\left\|\frac{X_{S^c}^\top r}{n\lambda}\right\|_\infty \le 1.
\end{align*}
[guided]
We first isolate the only optimality facts about the Lasso that are needed. Define the Lasso objective $Q_\lambda: \mathbb{R}^p \to \mathbb{R}$ by
\begin{align*}
Q_\lambda(b)
:=
\frac{1}{2n}\|Y-Xb\|_2^2+\lambda\|b\|_1.
\end{align*}
Fix one realization of the noise vector $\varepsilon$ and assume that, for this realization, a minimizer $\hat{\beta}$ has support $S$ and active sign vector $s$. Throughout this argument, $\operatorname{sign}$ is applied componentwise and $\operatorname{sign}(0)=0$. If $S=\varnothing$, the active-coordinate equations below are empty vector identities; the inactive-coordinate inequalities are still obtained coordinate by coordinate over $S^c$. Define the residual vector $r \in \mathbb{R}^n$ by
\begin{align*}
r := Y-X\hat{\beta}.
\end{align*}
For an active coordinate $j \in S$, the coefficient $\hat{\beta}_j$ is nonzero. This matters because the map $t \mapsto |\hat{\beta}_j+t|$ is differentiable at $t=0$, with derivative $\operatorname{sign}(\hat{\beta}_j)$. Let $e_j \in \mathbb{R}^p$ denote the $j$th standard basis vector. Define the one-variable map $\phi_j: \mathbb{R} \to \mathbb{R}$ by $\phi_j(t) := Q_\lambda(\hat{\beta}+t e_j)$. This map has a minimum at $t=0$. Since it is differentiable at $0$, its derivative at $0$ must be zero. Differentiating the quadratic term gives
\begin{align*}
\frac{d}{dt}\bigg|_{t=0}\frac{1}{2n}|Y-X(\hat{\beta}+t e_j)|^2
=
-\frac{1}{n}X_j^\top (Y-X\hat{\beta})
=
-\frac{1}{n}X_j^\top r.
\end{align*}
Differentiating the $\ell^1$ term in the active coordinate gives $\lambda \operatorname{sign}(\hat{\beta}_j)$. Hence
\begin{align*}
-\frac{1}{n}X_j^\top r+\lambda \operatorname{sign}(\hat{\beta}_j)=0.
\end{align*}
Collecting these equations for all $j \in S$ and using $\operatorname{sign}(\hat{\beta}_S)=s$ yields
\begin{align*}
\frac{X_S^\top r}{n}=\lambda s.
\end{align*}
For an inactive coordinate $j \in S^c$, the coefficient $\hat{\beta}_j$ equals $0$, so the absolute value is not differentiable at the origin. We therefore use one-sided perturbations instead of a derivative. For every $t \in \mathbb{R}$, minimality of $\hat{\beta}$ gives
\begin{align*}
0
\le
Q_\lambda(\hat{\beta}+t e_j)-Q_\lambda(\hat{\beta}).
\end{align*}
Since only the $j$th coefficient changes and $\hat{\beta}_j=0$, the $\ell^1$ term changes by $\lambda |t|$. The quadratic term expands as
\begin{align*}
\frac{1}{2n}|r-tX_j|^2-\frac{1}{2n}|r|^2
=
-\frac{t}{n}X_j^\top r+\frac{t^2}{2n}\|X_j\|_2^2.
\end{align*}
Therefore
\begin{align*}
0
\le
-\frac{t}{n}X_j^\top r
+
\frac{t^2}{2n}|X_j|^2
+
\lambda |t|.
\end{align*}
For $t>0$, dividing by $t$ gives
\begin{align*}
0
\le
-\frac{1}{n}X_j^\top r+\frac{t}{2n}\|X_j\|_2^2+\lambda.
\end{align*}
Letting $t \downarrow 0$ gives $-\frac{1}{n}X_j^\top r+\lambda \ge 0$, because the affine expression in $t$ on the right-hand side is continuous at $0$. For $t<0$, dividing by $t$ reverses the inequality:
\begin{align*}
0
\ge
-\frac{1}{n}X_j^\top r+\frac{t}{2n}\|X_j\|_2^2-\lambda.
\end{align*}
Letting $t \uparrow 0$ gives $-\frac{1}{n}X_j^\top r-\lambda \le 0$, again by continuity of the affine expression in $t$ at $0$. These two inequalities are exactly
\begin{align*}
-\lambda \le \frac{X_j^\top r}{n} \le \lambda,
\end{align*}
or
\begin{align*}
\left|\frac{X_j^\top r}{n\lambda}\right|\le 1.
\end{align*}
Taking the maximum over $j \in S^c$ gives the inactive coordinate condition
\begin{align*}
\left\|\frac{X_{S^c}^\top r}{n\lambda}\right\|_\infty \le 1.
\end{align*}
[/guided]
[/step]
[step:Solve the active equations and obtain the active sign barrier]
Since $\operatorname{supp}(\hat{\beta})=S$, we have $\hat{\beta}_{S^c}=0$. Using $Y=X\beta^*+\varepsilon$ and $\operatorname{supp}(\beta^*)=S$, the residual satisfies
\begin{align*}
r
=
Y-X\hat{\beta}
=
X_S(\beta^*_S-\hat{\beta}_S)+\varepsilon.
\end{align*}
Substituting this expression into the active optimality equation gives
\begin{align*}
\lambda s
=
\frac{X_S^\top r}{n}
=
\hat{\Sigma}_{SS}(\beta^*_S-\hat{\beta}_S)+\frac{X_S^\top\varepsilon}{n}.
\end{align*}
Since $\hat{\Sigma}_{SS}$ is invertible,
\begin{align*}
\hat{\beta}_S
=
\beta^*_S+\hat{\Sigma}_{SS}^{-1}\frac{X_S^\top\varepsilon}{n}
-\lambda \hat{\Sigma}_{SS}^{-1}s
=
\beta^*_S+W_S-\lambda \hat{\Sigma}_{SS}^{-1}s.
\end{align*}
The assumed equality $\operatorname{sign}(\hat{\beta}_S)=s$ therefore implies
\begin{align*}
\operatorname{sign}\left(\beta^*_S+W_S-\lambda\hat{\Sigma}_{SS}^{-1}s\right)=s.
\end{align*}
[guided]
The active equations identify what the Lasso coefficient on $S$ must be if exact signed recovery occurs. Since $\operatorname{supp}(\hat{\beta})=S$, we have $\hat{\beta}_{S^c}=0$. Since $\operatorname{supp}(\beta^*)=S$, the true signal satisfies $X\beta^*=X_S\beta^*_S$. Hence the residual vector is
\begin{align*}
r
=
Y-X\hat{\beta}
=
X_S\beta^*_S+\varepsilon-X_S\hat{\beta}_S
=
X_S(\beta^*_S-\hat{\beta}_S)+\varepsilon.
\end{align*}
The active optimality condition from the previous step is
\begin{align*}
\frac{X_S^\top r}{n}=\lambda s.
\end{align*}
Substituting the residual formula into this equation gives
\begin{align*}
\lambda s
=
\frac{X_S^\top}{n}\left(X_S(\beta^*_S-\hat{\beta}_S)+\varepsilon\right)
=
\hat{\Sigma}_{SS}(\beta^*_S-\hat{\beta}_S)+\frac{X_S^\top\varepsilon}{n}.
\end{align*}
Here we use the definition $\hat{\Sigma}_{SS}=X_S^\top X_S/n$. The hypothesis that $\hat{\Sigma}_{SS}$ is invertible is used exactly at this point: multiplying by $\hat{\Sigma}_{SS}^{-1}$ and solving for $\hat{\beta}_S$ gives
\begin{align*}
\hat{\beta}_S
=
\beta^*_S+\hat{\Sigma}_{SS}^{-1}\frac{X_S^\top\varepsilon}{n}
-\lambda\hat{\Sigma}_{SS}^{-1}s.
\end{align*}
By the definition of the active noise vector $W_S$, this is
\begin{align*}
\hat{\beta}_S=\beta^*_S+W_S-\lambda\hat{\Sigma}_{SS}^{-1}s.
\end{align*}
If the assumed minimizer has active sign vector $s$, then taking signs on both sides yields the necessary active barrier
\begin{align*}
\operatorname{sign}\left(\beta^*_S+W_S-\lambda\hat{\Sigma}_{SS}^{-1}s\right)=s.
\end{align*}
[/guided]
[/step]
[step:Substitute the active solution into the inactive dual condition]
Using the residual identity and the formula for $\hat{\beta}_S$, compute
\begin{align*}
\beta^*_S-\hat{\beta}_S
=
-\hat{\Sigma}_{SS}^{-1}\frac{X_S^\top\varepsilon}{n}
+
\lambda \hat{\Sigma}_{SS}^{-1}s.
\end{align*}
Substituting this expression into $X_{S^c}^\top r/(n\lambda)$ first gives
\begin{align*}
\frac{X_{S^c}^\top r}{n\lambda} = \frac{1}{\lambda}\left[\hat{\Sigma}_{S^cS}(\beta^*_S-\hat{\beta}_S)+\frac{X_{S^c}^\top\varepsilon}{n}\right].
\end{align*}
Using the displayed identity for $\beta^*_S-\hat{\beta}_S$, this becomes
\begin{align*}
\frac{X_{S^c}^\top r}{n\lambda} = \hat{\Sigma}_{S^cS}\hat{\Sigma}_{SS}^{-1}s + \frac{1}{\lambda}\left[\frac{X_{S^c}^\top\varepsilon}{n}-\hat{\Sigma}_{S^cS}\hat{\Sigma}_{SS}^{-1}\frac{X_S^\top\varepsilon}{n}\right].
\end{align*}
Let $I_n\in\mathbb{R}^{n\times n}$ denote the identity matrix. Since $\hat{\Sigma}_{S^cS}=X_{S^c}^\top X_S/n$, the bracketed noise term is
\begin{align*}
\frac{X_{S^c}^\top\varepsilon}{n}-\hat{\Sigma}_{S^cS}\hat{\Sigma}_{SS}^{-1}\frac{X_S^\top\varepsilon}{n} = \frac{X_{S^c}^\top}{n}\left(I_n-X_S\hat{\Sigma}_{SS}^{-1}\frac{X_S^\top}{n}\right)\varepsilon.
\end{align*}
By the definition of $R_{S^c}$, we therefore obtain
\begin{align*}
\frac{X_{S^c}^\top r}{n\lambda} = \hat{\Sigma}_{S^cS}\hat{\Sigma}_{SS}^{-1}s+\frac{1}{\lambda}R_{S^c}.
\end{align*}
Combining this identity with
\begin{align*}
\left\|\frac{X_{S^c}^\top r}{n\lambda}\right\|_\infty \le 1
\end{align*}
gives
\begin{align*}
\left\|
\hat{\Sigma}_{S^cS}\hat{\Sigma}_{SS}^{-1}s+\frac{1}{\lambda}R_{S^c}
\right\|_\infty \le 1.
\end{align*}
[guided]
The inactive condition starts from the bound already proved for every inactive coordinate:
\begin{align*}
\left\|\frac{X_{S^c}^\top r}{n\lambda}\right\|_\infty \le 1.
\end{align*}
We now rewrite the vector inside this norm in terms of deterministic design quantities and the residual noise. From the active formula,
\begin{align*}
\hat{\beta}_S
=
\beta^*_S+\hat{\Sigma}_{SS}^{-1}\frac{X_S^\top\varepsilon}{n}
-\lambda\hat{\Sigma}_{SS}^{-1}s,
\end{align*}
so
\begin{align*}
\beta^*_S-\hat{\beta}_S
=
-\hat{\Sigma}_{SS}^{-1}\frac{X_S^\top\varepsilon}{n}
+
\lambda\hat{\Sigma}_{SS}^{-1}s.
\end{align*}
Using $r=X_S(\beta^*_S-\hat{\beta}_S)+\varepsilon$, we obtain
\begin{align*}
\frac{X_{S^c}^\top r}{n\lambda}
=
\frac{1}{\lambda}\left[\hat{\Sigma}_{S^cS}(\beta^*_S-\hat{\beta}_S)+\frac{X_{S^c}^\top\varepsilon}{n}\right].
\end{align*}
Substituting the displayed expression for $\beta^*_S-\hat{\beta}_S$ gives
\begin{align*}
\frac{X_{S^c}^\top r}{n\lambda}
=
\hat{\Sigma}_{S^cS}\hat{\Sigma}_{SS}^{-1}s
+
\frac{1}{\lambda}\left[\frac{X_{S^c}^\top\varepsilon}{n}-\hat{\Sigma}_{S^cS}\hat{\Sigma}_{SS}^{-1}\frac{X_S^\top\varepsilon}{n}\right].
\end{align*}
The bracket is exactly the inactive residual-noise vector. Let $I_n\in\mathbb{R}^{n\times n}$ denote the identity matrix. Indeed, since $\hat{\Sigma}_{S^cS}=X_{S^c}^\top X_S/n$,
\begin{align*}
\frac{X_{S^c}^\top\varepsilon}{n}-\hat{\Sigma}_{S^cS}\hat{\Sigma}_{SS}^{-1}\frac{X_S^\top\varepsilon}{n}
=
\frac{X_{S^c}^\top}{n}\left(I_n-X_S\hat{\Sigma}_{SS}^{-1}\frac{X_S^\top}{n}\right)\varepsilon
=
R_{S^c}.
\end{align*}
Therefore
\begin{align*}
\frac{X_{S^c}^\top r}{n\lambda}
=
\hat{\Sigma}_{S^cS}\hat{\Sigma}_{SS}^{-1}s+\frac{1}{\lambda}R_{S^c}.
\end{align*}
Putting this identity into the inactive norm bound yields
\begin{align*}
\left\|
\hat{\Sigma}_{S^cS}\hat{\Sigma}_{SS}^{-1}s+\frac{1}{\lambda}R_{S^c}
\right\|_\infty \le 1.
\end{align*}
[/guided]
[/step]
[step:Convert the necessary conditions into the probability upper bound]
Let $A$ denote the event that there exists a Lasso minimizer $\hat{\beta}$ with
\begin{align*}
\operatorname{sign}(\hat{\beta})=\operatorname{sign}(\beta^*).
\end{align*}
On $A$, the convention $\operatorname{sign}(0)=0$ implies that the zero coordinates of $\hat{\beta}$ and $\beta^*$ coincide, so this minimizer satisfies $\operatorname{supp}(\hat{\beta})=S$ and $\operatorname{sign}(\hat{\beta}_S)=s$. The previous steps show that $A$ is contained in the event
\begin{align*}
B :=
\left\{
\operatorname{sign}\left(\beta^*_S+W_S-\lambda\hat{\Sigma}_{SS}^{-1}s\right)=s,
\,
\left\|
\hat{\Sigma}_{S^cS}\hat{\Sigma}_{SS}^{-1}s+\frac{1}{\lambda}R_{S^c}
\right\|_\infty \le 1
\right\}.
\end{align*}
Therefore monotonicity of probability gives
\begin{align*}
\mathbb{P}(A)\le \mathbb{P}(B),
\end{align*}
which is the stated probability inequality.
[guided]
We now turn the deterministic necessary conditions into a probability statement. Let $A$ be the event that there exists a Lasso minimizer $\hat{\beta}$ satisfying
\begin{align*}
\operatorname{sign}(\hat{\beta})=\operatorname{sign}(\beta^*).
\end{align*}
On this event, the convention $\operatorname{sign}(0)=0$ means that a coordinate is zero exactly when its sign is zero. Hence the support and sign assumptions used in the deterministic part are satisfied: $\operatorname{supp}(\hat{\beta})=S$ and $\operatorname{sign}(\hat{\beta}_S)=s$. Therefore every outcome in $A$ must also satisfy both barriers proved above. Define
\begin{align*}
B
:=
\left\{
\operatorname{sign}\left(\beta^*_S+W_S-\lambda\hat{\Sigma}_{SS}^{-1}s\right)=s,
\,
\left\|
\hat{\Sigma}_{S^cS}\hat{\Sigma}_{SS}^{-1}s+\frac{1}{\lambda}R_{S^c}
\right\|_\infty \le 1
\right\}.
\end{align*}
The previous steps prove the event inclusion $A\subseteq B$. This is the only probabilistic input: for any probability law of the noise vector $\varepsilon$, monotonicity of probability gives
\begin{align*}
\mathbb{P}(A)\le \mathbb{P}(B).
\end{align*}
Expanding the definitions of $A$ and $B$ is exactly the probability inequality stated in the theorem.
[/guided]
[/step]
[step:Use the deterministic incoherence violation and small residual noise to force failure]
For the sequence in the final assertion, write $\lambda_n>0$ for the Lasso tuning parameter at stage $n$, write $S_n$ for the true support at stage $n$, and write $s_n:=\operatorname{sign}(\beta^*_{n,S_n})$ for the active sign vector. Also write $\hat{\Sigma}_{n,S_n^cS_n}$ for the inactive-active design covariance matrix at stage $n$, write $\hat{\Sigma}_{n,S_nS_n}$ for the active-active design covariance matrix at stage $n$, and write $R_{n,S_n^c}$ for the inactive residual-noise vector at stage $n$. Define the deterministic inactive vector $a_n \in \mathbb{R}^{S_n^c}$ by
\begin{align*}
a_n
:=
\hat{\Sigma}_{n,S_n^cS_n}
\left(\hat{\Sigma}_{n,S_nS_n}\right)^{-1}s_n.
\end{align*}
and define the random inactive residual vector $u_n \in \mathbb{R}^{S_n^c}$ by
\begin{align*}
u_n := \frac{1}{\lambda_n}R_{n,S_n^c}.
\end{align*}
The elementary reverse triangle inequality for the supremum norm, obtained from $\|a_n\|_\infty\le \|a_n+u_n\|_\infty+\|u_n\|_\infty$, gives
\begin{align*}
\|a_n+u_n\|_\infty
\ge
\|a_n\|_\infty-\|u_n\|_\infty.
\end{align*}
On the event $\|R_{n,S_n^c}\|_\infty \le \lambda_n\delta$, we have $\|u_n\|_\infty \le \delta$. Therefore, for all sufficiently large $n$,
\begin{align*}
\|a_n+u_n\|_\infty
\ge
(1+2\delta)-\delta
=
1+\delta
>
1.
\end{align*}
Thus the inactive dual feasibility condition fails on the event $\|R_{n,S_n^c}\|_\infty \le \lambda_n\delta$. Since exact sign recovery implies inactive dual feasibility, the exact sign recovery event is contained in
\begin{align*}
\left\{\|R_{n,S_n^c}\|_\infty > \lambda_n\delta\right\}.
\end{align*}
Consequently,
\begin{align*}
\mathbb{P}(\text{exact Lasso sign recovery at stage } n)
\le
\mathbb{P}\left(\|R_{n,S_n^c}\|_\infty > \lambda_n\delta\right)
\to 0.
\end{align*}
This proves that the exact sign recovery probability tends to $0$ along the sequence.
[guided]
For the final assertion, we work stage by stage along the sequence. Write $\lambda_n>0$ for the Lasso tuning parameter at stage $n$, write $S_n$ for the true support at stage $n$, and write $s_n:=\operatorname{sign}(\beta^*_{n,S_n})$ for the active sign vector. Also write $\hat{\Sigma}_{n,S_n^cS_n}$ for the inactive-active design covariance matrix at stage $n$, write $\hat{\Sigma}_{n,S_nS_n}$ for the active-active design covariance matrix at stage $n$, and write $R_{n,S_n^c}$ for the inactive residual-noise vector at stage $n$. Define the deterministic inactive vector $a_n \in \mathbb{R}^{S_n^c}$ by
\begin{align*}
a_n
:=
\hat{\Sigma}_{n,S_n^cS_n}
\left(\hat{\Sigma}_{n,S_nS_n}\right)^{-1}s_n.
\end{align*}
and define the random inactive residual vector $u_n \in \mathbb{R}^{S_n^c}$ by
\begin{align*}
u_n:=\frac{1}{\lambda_n}R_{n,S_n^c}.
\end{align*}
The inactive dual feasibility condition at stage $n$ is $\|a_n+u_n\|_\infty\le 1$. The deterministic assumption says that
\begin{align*}
\|a_n\|_\infty\ge 1+2\delta.
\end{align*}
On the high-probability event $\|R_{n,S_n^c}\|_\infty\le \lambda_n\delta$, we also have $\|u_n\|_\infty\le\delta$. The elementary reverse triangle inequality for the supremum norm, obtained from $\|a_n\|_\infty\le \|a_n+u_n\|_\infty+\|u_n\|_\infty$, gives
\begin{align*}
\|a_n+u_n\|_\infty
\ge
\|a_n\|_\infty-\|u_n\|_\infty
\ge
(1+2\delta)-\delta
=
1+\delta
>
1.
\end{align*}
Thus inactive dual feasibility fails whenever $\|R_{n,S_n^c}\|_\infty\le \lambda_n\delta$. Since exact sign recovery implies inactive dual feasibility, the exact sign recovery event is contained in the complement event
\begin{align*}
\left\{\|R_{n,S_n^c}\|_\infty>\lambda_n\delta\right\}.
\end{align*}
Taking probabilities gives
\begin{align*}
\mathbb{P}(\text{exact Lasso sign recovery at stage } n)
\le
\mathbb{P}\left(\|R_{n,S_n^c}\|_\infty>\lambda_n\delta\right).
\end{align*}
The hypothesis $\mathbb{P}(\|R_{n,S_n^c}\|_\infty\le\lambda_n\delta)\to 1$ is equivalent to the right-hand side tending to $0$. Therefore the exact Lasso sign recovery probability tends to $0$ along the sequence.
[/guided]
[/step]
