From 6e70f28ccbd5afc1506f71f013278a9d157ef03a Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 27 Oct 2022 20:40:09 +0200
Subject: Optimize last images, add proof template, improve CSS

---
 source/know/concept/heaviside-step-function/index.md | 10 +++-------
 1 file changed, 3 insertions(+), 7 deletions(-)

(limited to 'source/know/concept/heaviside-step-function')

diff --git a/source/know/concept/heaviside-step-function/index.md b/source/know/concept/heaviside-step-function/index.md
index 15d1729..9f5d4ec 100644
--- a/source/know/concept/heaviside-step-function/index.md
+++ b/source/know/concept/heaviside-step-function/index.md
@@ -57,11 +57,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-fourier"/>
-<label for="proof-fourier">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-fourier">Proof.</label>
+
+{% include proof/start.html id="proof-fourier" -%}
 In this case, it is easiest to use $$\Theta(0) = 1/2$$,
 such that the Heaviside step function can be expressed
 using the signum function $$\mathrm{sgn}(t)$$:
@@ -88,9 +85,8 @@ $$\begin{aligned}
     &= A \pi \delta(s \omega) + \frac{A}{2} \pv{\int_{-\infty}^\infty \mathrm{sgn}(t) \exp(i s \omega t) \dd{t}}
     = \frac{A}{|s|} \pi \delta(\omega) + i \frac{A}{s} \pv{\frac{1}{\omega}}
 \end{aligned}$$
+{% include proof/end.html id="proof-fourier" %}
 
-</div>
-</div>
 
 The use of $$\pv{}$$ without an integral is an abuse of notation,
 and means that this result only makes sense when wrapped in an integral.
-- 
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