PomeProposition 17.10.43.61DD221912DE Let $\varphi ,\psi ,\gamma \in \mathcal{E}^1(X,\theta ;\phi )$. Then
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17.10.44
\begin{equation} d_1(\varphi ,\psi )\geq d_1(\varphi \land \gamma ,\psi \land \gamma ). \end{equation}
Proposition 17.10.4361DD221912DE
Proof. Step 1. We first assume that $\varphi \geq \psi $. Then
\[ \begin{aligned} d_1(\varphi ,\psi )=& E_{\theta }^{\phi }(\varphi )-E_{\theta }^{\phi }(\psi )\\ \geq & E_{\theta }^{\phi }\Bigl ((\varphi \land \gamma )\lor \psi \Bigr )-E_{\theta }^{\phi }(\psi ) & \textup{by \href{../../tag/93210731F7FD/}{16.5.14}}\\ \geq & E_{\theta }^{\phi }(\varphi \land \gamma )-E_{\theta }^{\phi }(\psi \land \gamma )& \textup{by corollary \href{../../tag/4A849C02F386/}{16.5.59}}. \end{aligned} \]
Step 2. We prove the general case.
By Step 1, we have
\[ d_1(\varphi ,\varphi \land \psi )\geq d_1(\varphi \land \gamma ,\varphi \land \psi \land \gamma ),\quad d_1(\psi ,\varphi \land \psi )\geq d_1(\psi \land \gamma ,\varphi \land \psi \land \gamma ). \]
Adding the two inequalities together, we conclude (17.10.44). $\square$
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