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/impulse-response/index.md | 10 +++-------
 1 file changed, 3 insertions(+), 7 deletions(-)

(limited to 'source/know/concept/impulse-response')

diff --git a/source/know/concept/impulse-response/index.md b/source/know/concept/impulse-response/index.md
index 397ac2d..661ed3f 100644
--- a/source/know/concept/impulse-response/index.md
+++ b/source/know/concept/impulse-response/index.md
@@ -30,11 +30,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-main"/>
-<label for="proof-main">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-main">Proof.</label>
+
+{% include proof/start.html id="proof-theorem" -%}
 Starting from the definition of $$u_p(t)$$,
 we shift the argument by some constant $$\tau$$,
 and multiply both sides by the constant $$f(\tau)$$:
@@ -60,9 +57,8 @@ $$\begin{aligned}
     \hat{L} \int_0^\infty f(\tau) \: u_p(t - \tau) \dd{\tau}
     &= (f * u_p)(t) = \hat{L}\{ u(t) \} = f(t)
 \end{aligned}$$
+{% include proof/end.html id="proof-theorem" %}
 
-</div>
-</div>
 
 This is useful for solving initial value problems,
 because any initial condition can be satisfied
-- 
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