Optimize last images, add proof template, improve CSS

30 files changed, 234 insertions, 527 deletions

diff --git a/source/know/concept/binomial-distribution/index.md b/source/know/concept/binomial-distribution/index.md
index 1193a93..dc75221 100644
--- a/source/know/concept/binomial-distribution/index.md
+++ b/source/know/concept/binomial-distribution/index.md
@@ -44,11 +44,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-mean"/>
-<label for="proof-mean">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-mean">Proof.</label>
+
+{% include proof/start.html id="proof-mean" -%}
 The trick is to treat $$p$$ and $$q$$ as independent until the last moment:
 
 $$\begin{aligned}
@@ -62,8 +59,8 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Inserting $$q = 1 - p$$ then gives the desired result.
-</div>
-</div>
+{% include proof/end.html id="proof-mean" %}
+
 
 Meanwhile, we find the following variance $$\sigma^2$$,
 with $$\sigma$$ being the standard deviation:
@@ -74,12 +71,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-var"/>
-<label for="proof-var">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-var">Proof.</label>
-We use the same trick to calculate $$\overline{n^2}$$
+
+{% include proof/start.html id="proof-var" -%}
 (the mean squared number of successes):
 
 $$\begin{aligned}
@@ -106,8 +99,8 @@ $$\begin{aligned}
 \end{aligned}$$
 
 By inserting $$q = 1 - p$$, we arrive at the desired expression.
-</div>
-</div>
+{% include proof/end.html id="proof-var" %}
+
 
 As $$N \to \infty$$, the binomial distribution
 turns into the continuous normal distribution,
@@ -119,11 +112,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-normal"/>
-<label for="proof-normal">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-normal">Proof.</label>
+
+{% include proof/start.html id="proof-normal" -%}
 We take the Taylor expansion of $$\ln\!\big(P_N(n)\big)$$
 around the mean $$\mu = Np$$:
 
@@ -211,8 +201,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Taking $$\exp$$ of this expression then yields a normalized Gaussian distribution.
-</div>
-</div>
+{% include proof/end.html id="proof-normal" %}
 
 
 ## References
diff --git a/source/know/concept/boltzmann-equation/index.md b/source/know/concept/boltzmann-equation/index.md
index 9ed2fd2..d2631b2 100644
--- a/source/know/concept/boltzmann-equation/index.md
+++ b/source/know/concept/boltzmann-equation/index.md
@@ -145,11 +145,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-moment0"/>
-<label for="proof-moment0">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-moment0">Proof.</label>
+
+{% include proof/start.html id="proof-moment0" -%}
 We insert $$Q = m$$ into our prototype,
 and since $$m$$ is constant, the rest is trivial:
 
@@ -159,9 +156,8 @@ $$\begin{aligned}
     \\
     &= \pdv{\rho}{t} + \nabla \cdot \big(\rho \Expval{\vb{v}}\big) - 0
 \end{aligned}$$
+{% include proof/end.html id="proof-moment0" %}
 
-</div>
-</div>
 
 If we instead choose the momentum $$Q = m \vb{v}$$,
 we find that the **first moment** of the BTE describes conservation of momentum,
@@ -174,11 +170,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-moment1"/>
-<label for="proof-moment1">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-moment1">Proof.</label>
+
+{% include proof/start.html id="proof-moment1" -%}
 We insert $$Q = m \vb{v}$$ into our prototype and recognize $$\rho$$ wherever possible:
 
 $$\begin{aligned}
@@ -220,9 +213,8 @@ $$\begin{aligned}
     0
     &= \pdv{}{t}\big(\rho \vb{V}\big) + \nabla \cdot \big(\rho \vb{V} \vb{V} + \hat{P}\big) - n \vb{F}
 \end{aligned}$$
+{% include proof/end.html id="proof-moment1" %}
 
-</div>
-</div>
 
 Finally, if we choose the kinetic energy $$Q = m |\vb{v}|^2 / 2$$,
 we find that the **second moment** gives conservation of energy,
@@ -237,11 +229,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-moment2"/>
-<label for="proof-moment2">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-moment2">Proof.</label>
+
+{% include proof/start.html id="proof-moment2" -%}
 We insert $$Q = m |\vb{v}|^2 / 2$$ into our prototype and recognize $$\rho$$ wherever possible:
 
 $$\begin{aligned}
@@ -349,9 +338,7 @@ $$\begin{aligned}
     \end{bmatrix}
     = \sum_{i=1}^{3} \sum_{j=1}^{3} \pdv{P_{ij}}{x_j} V_i
 \end{aligned}$$
-
-</div>
-</div>
+{% include proof/end.html id="proof-moment2" %}
 
 
 
diff --git a/source/know/concept/convolution-theorem/index.md b/source/know/concept/convolution-theorem/index.md
index 742c8ff..510417a 100644
--- a/source/know/concept/convolution-theorem/index.md
+++ b/source/know/concept/convolution-theorem/index.md
@@ -12,6 +12,8 @@ is equal to a product in the frequency domain. This is especially useful
 for computation, replacing an $$\mathcal{O}(n^2)$$ convolution with an
 $$\mathcal{O}(n \log(n))$$ transform and product.
 
+
+
 ## Fourier transform
 
 The convolution theorem is usually expressed as follows, where
@@ -27,11 +29,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<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" -%}
 We expand the right-hand side of the theorem and
 rearrange the integrals:
 
@@ -57,8 +56,8 @@ $$\begin{aligned}
     &= B \int_{-\infty}^\infty \tilde{g}(k') \: \tilde{f}(k - k') \dd{k'}
     = B \cdot (\tilde{f} * \tilde{g})(k)
 \end{aligned}$$
-</div>
-</div>
+{% include proof/end.html id="proof-fourier" %}
+
 
 
 ## Laplace transform
@@ -79,11 +78,8 @@ $$\begin{aligned}
     \boxed{\hat{\mathcal{L}}\{(f * g)(t)\} = \tilde{f}(s) \: \tilde{g}(s)}
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-laplace"/>
-<label for="proof-laplace">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-laplace">Proof.</label>
+
+{% include proof/start.html id="proof-laplace" -%}
 We expand the left-hand side.
 Note that the lower integration limit is 0 instead of $$-\infty$$,
 because we set both $$f(t)$$ and $$g(t)$$ to zero for $$t < 0$$:
@@ -106,8 +102,7 @@ $$\begin{aligned}
     &= \int_0^\infty \tilde{f}(s) \: g(t') \exp(- s t') \dd{t'}
     = \tilde{f}(s) \: \tilde{g}(s)
 \end{aligned}$$
-</div>
-</div>
+{% include proof/end.html id="proof-laplace" %}
 
 
 
diff --git a/source/know/concept/curvilinear-coordinates/index.md b/source/know/concept/curvilinear-coordinates/index.md
index cb22e43..48a5a72 100644
--- a/source/know/concept/curvilinear-coordinates/index.md
+++ b/source/know/concept/curvilinear-coordinates/index.md
@@ -48,6 +48,7 @@ we derive general formulae to convert expressions
 from Cartesian coordinates to the new orthogonal system $$(x_1, x_2, x_3)$$.
 
 
+
 ## Basis vectors
 
 Consider the the vector form of the line element $$\dd{\ell}$$,
@@ -86,6 +87,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 
+
 ## Gradient
 
 In an orthogonal coordinate system,
@@ -102,11 +104,8 @@ $$\begin{gathered}
     }
 \end{gathered}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-grad"/>
-<label for="proof-grad">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-grad">Proof.</label>
+
+{% include proof/start.html id="proof-grad" -%}
 For a direction $$\dd{\ell}$$, we know that
 $$\idv{f}{\ell}$$ is the component of $$\nabla f$$ in that direction:
 
@@ -127,9 +126,8 @@ $$\begin{gathered}
     + \vu{e}_2 \dv{x_2}{\ell} \pdv{f}{x_2}
     + \vu{e}_3 \dv{x_3}{\ell} \pdv{f}{x_3}
 \end{gathered}$$
+{% include proof/end.html id="proof-grad" %}
 
-</div>
-</div>
 
 
 ## Divergence
@@ -145,11 +143,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-div"/>
-<label for="proof-div">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-div">Proof.</label>
+
+{% include proof/start.html id="proof-div" -%}
 As preparation, we rewrite $$\vb{V}$$ as follows
 to introduce the scale factors:
 
@@ -222,8 +217,8 @@ $$\begin{aligned}
 
 After repeating this procedure for the other components of $$\vb{V}$$,
 we get the desired general expression for the divergence.
-</div>
-</div>
+{% include proof/end.html id="proof-div" %}
+
 
 
 ## Laplacian
@@ -246,6 +241,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 
+
 ## Curl
 
 The curl of a vector $$\vb{V}$$ is as follows
@@ -264,11 +260,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-curl"/>
-<label for="proof-curl">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-curl">Proof.</label>
+
+{% include proof/start.html id="proof-curl" -%}
 The curl is found in a similar way as the divergence.
 We rewrite $$\vb{V}$$ like so:
 
@@ -317,8 +310,8 @@ $$\begin{aligned}
 
 If we go through the same process for the other components of $$\vb{V}$$
 and add up the results, we get the desired expression for the curl.
-</div>
-</div>
+{% include proof/end.html id="proof-curl" %}
+
 
 
 ## Differential elements
diff --git a/source/know/concept/detailed-balance/index.md b/source/know/concept/detailed-balance/index.md
index b89d5da..98f9bd3 100644
--- a/source/know/concept/detailed-balance/index.md
+++ b/source/know/concept/detailed-balance/index.md
@@ -103,11 +103,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-reversibility"/>
-<label for="proof-reversibility">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-reversibility">Proof.</label>
+
+{% include proof/start.html id="proof-reversibility" -%}
 Consider the following weighted inner product,
 whose weight function is a stationary distribution $$\pi$$
 satisfying detailed balance,
@@ -222,8 +219,7 @@ $$\begin{aligned}
 
 Where the integral gave the expectation value at $$X_0$$,
 since $$\pi$$ does not change in time.
-</div>
-</div>
+{% include proof/end.html id="proof-reversibility" %}
 
 
 
diff --git a/source/know/concept/dirac-delta-function/index.md b/source/know/concept/dirac-delta-function/index.md
index 518eba1..0185b78 100644
--- a/source/know/concept/dirac-delta-function/index.md
+++ b/source/know/concept/dirac-delta-function/index.md
@@ -65,11 +65,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-scale"/>
-<label for="proof-scale">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-scale">Proof.</label>
+
+{% include proof/start.html id="proof-scale" -%}
 Because it is symmetric, $$\delta(s x) = \delta(|s| x)$$.
 Then by substituting $$\sigma = |s| x$$:
 
@@ -77,9 +74,8 @@ $$\begin{aligned}
     \int \delta(|s| x) \dd{x}
     &= \frac{1}{|s|} \int \delta(\sigma) \dd{\sigma} = \frac{1}{|s|}
 \end{aligned}$$
+{% include proof/end.html id="proof-scale" %}
 
-</div>
-</div>
 
 An even more impressive property is the behaviour of the derivative of $$\delta(x)$$:
 
@@ -89,11 +85,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-dv1"/>
-<label for="proof-dv1">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-dv1">Proof.</label>
+
+{% include proof/start.html id="proof-dv1" -%}
 Note which variable is used for the
 differentiation, and that $$\delta'(x - \xi) = - \delta'(\xi - x)$$:
 
@@ -102,9 +95,8 @@ $$\begin{aligned}
     &= \dv{}{x}\int f(\xi) \: \delta(x - \xi) \dd{x}
     = f'(x)
 \end{aligned}$$
+{% include proof/end.html id="proof-dv1" %}
 
-</div>
-</div>
 
 This property also generalizes nicely for the higher-order derivatives:
 
diff --git a/source/know/concept/dynkins-formula/index.md b/source/know/concept/dynkins-formula/index.md
index c0d20c5..307f098 100644
--- a/source/know/concept/dynkins-formula/index.md
+++ b/source/know/concept/dynkins-formula/index.md
@@ -39,11 +39,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-kolmogorov"/>
-<label for="proof-kolmogorov">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-kolmogorov">Proof.</label>
+
+{% include proof/start.html id="proof-kolmogorov" -%}
 We define a new process $$Y_t \equiv h(X_t)$$, and then apply Itō's lemma, leading to:
 
 $$\begin{aligned}
@@ -84,9 +81,8 @@ $$\begin{aligned}
     \hat{L}\{h(X_0)\}
     \approx \frac{1}{t} \mathbf{E}[Y_t - Y_0| X_0]
 \end{aligned}$$
+{% include proof/end.html id="proof-kolmogorov" %}
 
-</div>
-</div>
 
 The general definition of resembles that of a classical derivative,
 and indeed, the generator $$\hat{A}$$ can be thought of as a differential operator.
@@ -104,11 +100,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-dynkin"/>
-<label for="proof-dynkin">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-dynkin">Proof.</label>
+
+{% include proof/start.html id="proof-dynkin" -%}
 The proof is similar to the one above.
 Define $$Y_t = h(X_t)$$ and use Itō’s lemma:
 
@@ -136,9 +129,9 @@ $$\begin{aligned}
     = \mathbf{E}\bigg[ Y_\tau - Y_0 - \int_0^\tau \hat{L}\{h(X_t)\} \dd{t} \bigg| X_0 \bigg]
 \end{aligned}$$
 
-Isolating this equation for $$\mathbf{E}[Y_\tau | X_0]$$ then gives Dynkin's formula.
-</div>
-</div>
+Isolating this equation for $$\mathbf{E}[Y_\tau \!\mid\! X_0]$$ then gives Dynkin's formula.
+{% include proof/end.html id="proof-dynkin" %}
+
 
 A common application of Dynkin's formula is predicting
 when the stopping time $$\tau$$ occurs, and in what state $$X_\tau$$ this happens.
diff --git a/source/know/concept/equation-of-motion-theory/index.md b/source/know/concept/equation-of-motion-theory/index.md
index 02ed856..c1ed8da 100644
--- a/source/know/concept/equation-of-motion-theory/index.md
+++ b/source/know/concept/equation-of-motion-theory/index.md
@@ -63,11 +63,8 @@ $$\begin{aligned}
     = - \sum_{\nu''} u_{\nu \nu''} \hat{c}_{\nu''}
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-commH0"/>
-<label for="proof-commH0">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-commH0">Proof.</label>
+
+{% include proof/start.html id="proof-commutator" -%}
 Using the commutator identity for $$\comm{A B}{C}$$,
 we decompose it like so:
 
@@ -105,9 +102,8 @@ $$\begin{aligned}
     - 2 \acomm{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''} \Big)
     = - \sum_{\nu''} u_{\nu \nu''} \hat{f}_{\!\nu''}
 \end{aligned}$$
+{% include proof/end.html id="proof-commutator" %}
 
-</div>
-</div>
 
 Substituting this into $$G_{\nu \nu'}^R$$'s equation of motion,
 we recognize another Green's function $$G_{\nu'' \nu'}^R$$:
diff --git a/source/know/concept/euler-bernoulli-law/index.md b/source/know/concept/euler-bernoulli-law/index.md
index dad67ca..5a6c38d 100644
--- a/source/know/concept/euler-bernoulli-law/index.md
+++ b/source/know/concept/euler-bernoulli-law/index.md
@@ -81,11 +81,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-field"/>
-<label for="proof-field">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-field">Proof.</label>
+
+{% include proof/start.html id="proof-field" -%}
 By integrating the above strains $$u_{ii} = \ipdv{u_i}{i}$$,
 we get the components of $$\va{u}$$:
 
@@ -171,8 +168,8 @@ $$\begin{aligned}
 
 Inserting this into the components $$u_x$$, $$u_y$$ and $$u_z$$
 then yields the full displacement field.
-</div>
-</div>
+{% include proof/end.html id="proof-field" %}
+
 
 In any case, the beam experiences a bending torque with an $$x$$-component $$T_x$$ given by:
 
diff --git a/source/know/concept/fourier-transform/index.md b/source/know/concept/fourier-transform/index.md
index 0bc849b..c86d997 100644
--- a/source/know/concept/fourier-transform/index.md
+++ b/source/know/concept/fourier-transform/index.md
@@ -67,6 +67,7 @@ on whether the analysis is for forward ($$s > 0$$) or backward-propagating
 ($$s < 0$$) waves.
 
 
+
 ## Derivatives
 
 The FT of a derivative has a very useful property.
@@ -113,6 +114,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 
+
 ## Multiple dimensions
 
 The Fourier transform is straightforward to generalize to $$N$$ dimensions.
@@ -150,11 +152,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-constants-ND"/>
-<label for="proof-constants-ND">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-constants-ND">Proof.</label>
+
+{% include proof/start.html id="proof-constants-ndim" -%}
 The inverse FT of the forward FT of $$f(\vb{x})$$ must be equal to $$f(\vb{x})$$ again, so:
 
 $$\begin{aligned}
@@ -180,9 +179,8 @@ $$\begin{aligned}
     &= \frac{(2 \pi)^N A B}{|s|^N} \int f(\vb{x}') \: \delta(\vb{x}' - \vb{x}) \ddn{N}{\vb{x}'}
     = \frac{(2 \pi)^N A B}{|s|^N} f(\vb{x})
 \end{aligned}$$
+{% include proof/end.html id="proof-constants-ndim" %}
 
-</div>
-</div>
 
 Differentiation is more complicated for $$N > 1$$,
 but the FT is still useful,
@@ -197,11 +195,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-laplacian"/>
-<label for="proof-laplacian">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-laplacian">Proof.</label>
+
+{% include proof/start.html id="proof-laplacian" -%}
 We insert $$\nabla^2 f$$ into the FT,
 decompose the exponential and the Laplacian,
 and then integrate by parts (limits $$\pm \infty$$ omitted):
@@ -236,9 +231,7 @@ $$\begin{aligned}
     &= - A s^2 \sum_{n = 1}^N k_n^2 \int f \exp(i s \vb{k} \cdot \vb{x}) \ddn{N}{\vb{x}}
     = - s^2 \sum_{n = 1}^N k_n^2 \tilde{f}
 \end{aligned}$$
-
-</div>
-</div>
+{% include proof/end.html id="proof-laplacian" %}
 
 
 
diff --git a/source/know/concept/fundamental-solution/index.md b/source/know/concept/fundamental-solution/index.md
index 312cc2e..947aada 100644
--- a/source/know/concept/fundamental-solution/index.md
+++ b/source/know/concept/fundamental-solution/index.md
@@ -42,11 +42,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-solution"/>
-<label for="proof-solution">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-solution">Proof.</label>
+
+{% include proof/start.html id="proof-solution" -%}
 $$\hat{L}$$ only acts on $$x$$, so $$x' \in ]a, b[$$ is simply a parameter,
 meaning we are free to multiply the definition of $$G$$
 by the constant $$f(x')$$ on both sides,
@@ -72,8 +69,8 @@ $$\begin{aligned}
 
 By definition, $$\hat{L}$$'s response $$u(x)$$ to $$f(x)$$
 satisfies $$\hat{L}\{ u(x) \} = f(x)$$, recognizable here.
-</div>
-</div>
+{% include proof/end.html id="proof-solution" %}
+
 
 While the impulse response is typically used for initial value problems,
 the fundamental solution $$G$$ is used for boundary value problems.
@@ -117,11 +114,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-reciprocity"/>
-<label for="proof-reciprocity">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-reciprocity">Proof.</label>
+
+{% include proof/start.html id="proof-reciprocity" -%}
 Consider two parameters $$x_1'$$ and $$x_2'$$.
 The self-adjointness of $$\hat{L}$$ means that:
 
@@ -135,9 +129,7 @@ $$\begin{aligned}
     G^*(x_2', x_1')
     &= G(x_1', x_2')
 \end{aligned}$$
-
-</div>
-</div>
+{% include proof/end.html id="proof-reciprocity" %}
 
 
 
diff --git a/source/know/concept/greens-functions/index.md b/source/know/concept/greens-functions/index.md
index ddba2cd..eda5671 100644
--- a/source/know/concept/greens-functions/index.md
+++ b/source/know/concept/greens-functions/index.md
@@ -21,6 +21,7 @@ but in general they are not the same,
 except in a special case, see below.
 
 
+
 ## Single-particle functions
 
 If the two operators are single-particle creation/annihilation operators,
@@ -146,11 +147,8 @@ $$\begin{gathered}
     G_{\nu \nu'}^<(t, t') = G_{\nu \nu'}^<(t - t')
 \end{gathered}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-time-diff"/>
-<label for="proof-time-diff">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-time-diff">Proof.</label>
+
+{% include proof/start.html id="proof-time-delta" -%}
 We will prove that the thermal expectation value
 $$\expval{\hat{A}(t) \hat{B}(t')}$$ only depends on $$t - t'$$
 for arbitrary $$\hat{A}$$ and $$\hat{B}$$,
@@ -189,8 +187,7 @@ because $$\hat{H}$$ is time-independent by assumption.
 Note that thermodynamic equilibrium is crucial:
 intuitively, if the system is not in equilibrium,
 then it evolves in some transient time-dependent way.
-</div>
-</div>
+{% include proof/end.html id="proof-time-delta" %}
 
 If the Hamiltonian is both time-independent and non-interacting,
 then the time-dependence of $$\hat{c}_\nu$$
@@ -214,6 +211,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 
+
 ## As fundamental solutions
 
 In the absence of interactions,
@@ -237,11 +235,8 @@ $$\begin{aligned}
     = \frac{\hbar^2}{2 m} \nabla^2 \hat{\Psi}(\vb{r})
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-commH0"/>
-<label for="proof-commH0">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-commH0">Proof.</label>
+
+{% include proof/start.html id="proof-commutator" -%}
 In the second quantization,
 the Hamiltonian $$\hat{H}_0$$ is written like so:
 
@@ -307,9 +302,8 @@ $$\begin{aligned}
     &= \frac{\hbar^2}{2 m} \sum_{\nu'} \hat{c}_{\nu'} \nabla^2 \psi_{\nu'}(\vb{r})
     = \frac{\hbar^2}{2 m} \nabla^2 \hat{\Psi}(\vb{r})
 \end{aligned}$$
+{% include proof/end.html id="proof-commutator" %}
 
-</div>
-</div>
 
 After substituting this into the equation of motion,
 we recognize $$G^R(\vb{r}, t; \vb{r}', t')$$ itself:
diff --git a/source/know/concept/gronwall-bellman-inequality/index.md b/source/know/concept/gronwall-bellman-inequality/index.md
index 8096aaf..da1bcad 100644
--- a/source/know/concept/gronwall-bellman-inequality/index.md
+++ b/source/know/concept/gronwall-bellman-inequality/index.md
@@ -26,11 +26,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-original"/>
-<label for="proof-original">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-original">Proof.</label>
+
+{% include proof/start.html id="proof-original" -%}
 We define $$w(t)$$ to equal the upper bounds above
 on both $$w'(t)$$ and $$w(t)$$ itself:
 
@@ -63,8 +60,8 @@ $$\begin{aligned}
 
 Since $$u' \le \beta u$$ as a condition,
 the above derivative is always negative.
-</div>
-</div>
+{% include proof/end.html id="proof-original" %}
+
 
 Grönwall's inequality can be generalized to non-differentiable functions.
 Suppose we know:
@@ -84,11 +81,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-integral"/>
-<label for="proof-integral">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-integral">Proof.</label>
+
+{% include proof/start.html id="proof-integral" -%}
 We start by defining $$w(t)$$ as follows,
 which will act as shorthand:
 
@@ -138,8 +132,8 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Insert this into the condition under which the Grönwall-Bellman inequality holds.
-</div>
-</div>