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\newcommand{\empty}{\varnothing}  \newcommand{\wp}[2]{wp\left(#1, #2\right)}  \newcommand{\if}[2]{if\left(#1\right) \ \{ #2 \}}  \newcommand{\else}[1]{else \ \{ #1 \}}  \newcommand{\ie}[3]{\if{#1}{#2} \ \else{#3}}  \newcommand{\a}[0]{\ \land \ }  \newcommand{\mtt}[1]{\text{#1}}  \newcommand{\mbb}[1]{\mathbf{#1}}  \newcommand{\T}[0]{\mbb{\top}}  \newcommand{\F}[0]{\mbb{\bot}}  \newcommand{\rule}[1]{\ (\mbb{§#1})}  \subsection{Variant decreases}  Prove that $I \a B \to \wp{V_1 := V; S}{V < V_1}$.  \begin{equation}  \begin{split}  & I \a B \to \wp{V_1 := V; S}{V < V_1}\\  \end{split}  \end{equation}  It is trivially true since $\forall n \in \mathbb{Z} : n - 1 < n = \T$.  It is easy to see that the variant decreases.