Variants of Farkas' Lemma
Let $A\in\mathbb{R}^{m\times n}$ and $b\in\mathbb{R}^m$. The following two variants of Farkas' lemma are equivalent:
\begin{align*}
&\exists y\in\mathbb{R}^m:\ A^Ty\geq 0,\ b^Ty<0 &&\overset{(F)}{\nLeftrightarrow} &&\exists x\in\mathbb{R}^n:\ Ax=b,\ x\geq 0\\
&\exists y\in\mathbb{R}^m:\ A^Ty=0,\ y\geq 0,\ b^Ty<0&&\overset{(V)}{\nLeftrightarrow} &&\exists x \in \mathbb{R}^n:\ Ax \leq b
\end{align*}
$(F)\Rightarrow(V)$
\begin{align*}
&\exists y \in \mathbb{R}^m: &&A^Ty=0,\ y\geq 0, &&b^Ty<0\\
\Leftrightarrow\ &\exists y \in \mathbb{R}^m: &&
\begin{pmatrix}
-A^T\\
A^T\\
E
\end{pmatrix}
y \geq 0, &&b^Ty<0\\
\overset{(F)}{\nLeftrightarrow}\ &\exists
\begin{pmatrix}
\hat{x}\\
\check{x}\\
\tilde{x}
\end{pmatrix}
\in \mathbb{R}^{2n+m}: &&
\begin{pmatrix}
-A^T\\
A^T\\
E
\end{pmatrix}^T
\begin{pmatrix}
\hat{x}\\
\check{x}\\
\tilde{x}
\end{pmatrix}
= b,
&&
\begin{pmatrix}
\hat{x}\\
\check{x}\\
\tilde{x}
\end{pmatrix}
\geq 0\\
\Leftrightarrow\ &\hat{x},\check{x}\in\mathbb{R}^n, \tilde{x}\in\mathbb{R}^m: &&-A\hat{x} + A\check{x} + \tilde{x} = b &&\hat{x},\check{x},\tilde{x}\geq 0\\
\Leftrightarrow\ &\hat{x},\check{x}\in\mathbb{R}^n: &&-A\hat{x} + A\check{x} \leq b &&\hat{x},\check{x}\geq 0\\
\Leftrightarrow\ &\hat{x},\check{x}\in\mathbb{R}^n: &&A\left(\check{x} - \hat{x}\right) \leq b &&\hat{x},\check{x}\geq 0\\
\overset{(*)}{\Leftrightarrow}\ &\exists x \in \mathbb{R}^n: &&Ax = b
\end{align*}
$\overset{(*)}{\Leftarrow}$: For $x\in\mathbb{R}^n$ with $Ax=b$ choose $\check{x},\hat{x}\in\mathbb{R}^n$ with:
\begin{align*}
&\check{x}_i :=
\begin{cases}
x_i, &x_i\geq0\\
0, &x_i < 0
\end{cases}
&&
\hat{x}_i :=
\begin{cases}
0, &x_i\geq0\\
-x_i, &x_i < 0
\end{cases}
\end{align*}
Then we have $\check{x}, \hat{x} \geq 0$ as well as $x=\check{x}-\hat{x}$ and therefore $A(\check{x} - \hat{x}) = b$.
$(F)\Leftarrow(V)$
\begin{align*}
&\exists x\in\mathbb{R}^n: &&Ax=b, &&x\geq 0\\
\Leftrightarrow\ &\exists x\in\mathbb{R}^n: &&Ax=b, &&-x\leq 0\\
\Leftrightarrow\ &\exists x\in\mathbb{R}^n: &&
\begin{pmatrix}
A\\
-A\\
-E
\end{pmatrix}
x \leq
\begin{pmatrix}
b\\
-b\\
0
\end{pmatrix}\\
\overset{(L)}{\nLeftrightarrow}\ &\exists
\begin{pmatrix}
\hat{y}\\
\check{y}\\
\tilde{y}
\end{pmatrix}
\in\mathbb{R}^{2m+n}:
&&
\begin{pmatrix}
A\\
-A\\
-E
\end{pmatrix}^T
\begin{pmatrix}
\hat{y}\\
\check{y}\\
\tilde{y}
\end{pmatrix}
=0,\
\begin{pmatrix}
\hat{y}\\
\check{y}\\
\tilde{y}
\end{pmatrix}
\geq 0,
&&
\begin{pmatrix}
b\\
-b\\
0
\end{pmatrix}^T
\begin{pmatrix}
\hat{y}\\
\check{y}\\
\tilde{y}
\end{pmatrix}
< 0\\
\Leftrightarrow\ &\exists \hat{y},\check{y}\in\mathbb{R}^{m}:
&&A^T\hat{y} - A^T\check{y} = 0,\ \hat{y},\check{y}\geq 0, &&b^T\hat{y}-b^T\check{y}<0\\
\Leftrightarrow\ &\exists \hat{y},\check{y}\in\mathbb{R}^{m}:
&&A^T\left(\hat{y} -\check{y}\right) = 0,\ \hat{y},\check{y}\geq 0, &&b^T\left(\hat{y}-\check{y}\right)<0\\
\overset{(*)}{\Leftrightarrow}\ &\exists y\in\mathbb{R}^{m}:
&&A^Ty = 0,\ y\geq 0, &&b^Ty<0
\end{align*}
$\overset{(*)}{\Leftarrow}$: Analogous.
Note
Your variant of Farkas' lemma looked like this:
\begin{align*}
&\exists y'\in\mathbb{R}^m:\ A^Ty'=0,\ y'\geq 0,\ b^Ty'=-1&&\overset{(V')}{\nLeftrightarrow} &&\exists x \in \mathbb{R}^n:\ Ax \leq b
\end{align*}
However, $(V)$ and $(V')$ are equivalent for the following reason:
\begin{align*}
&\exists y\in\mathbb{R}^m: &&A^Ty=0, &&y\geq 0, &&b^Ty<0\\
\overset{(*)}{\Leftrightarrow}\ &\exists y'\in\mathbb{R}^m: &&A^Ty'=0, &&y'\geq 0, &&b^Ty'=-1
\end{align*}
$\overset{(*)}{\Rightarrow}$: Choose $y' := -\left(b^Ty\right)^{-1} y \in \mathbb{R}^m$ for $y \in \mathbb{R}^m$ with $A^Ty=0$, $y\geq 0$ and $b^Ty<0$. Then we have $A^Ty'=0$, $y'\geq 0$ and $b^Ty'=-1$.
