To prove the first assertion we have to prove that MS1, MS2, MS3 and MS5, MS6 hold. \medskip\noindent Proof of MS1. It is clear that identities are in $S$ because $(X, X, 0, 1, 0, 0)$ is distinguished for every object $X$ of $\mathcal{D}$ and because $0$ is an object of $\mathcal{D}'$. Let $f : X \to Y$ and $g : Y \to Z$ be composable morphisms contained in $S$. Choose distinguished triangles $(X, Y, Q_1, f, p_1, d_1)$, $(X, Z, Q_2, g \circ f, p_2, d_2)$, and $(Y, Z, Q_3, g, p_3, d_3)$. By assumption we know that $Q_1$ and $Q_3$ are isomorphic to objects of $\mathcal{D}'$. By TR4 we know there exists a distinguished triangle $(Q_1, Q_2, Q_3, a, b, c)$. Since $\mathcal{D}'$ is a triangulated subcategory we conclude that $Q_2$ is isomorphic to an object of $\mathcal{D}'$. Hence $g \circ f \in S$. \medskip\noindent Proof of MS3. Let $a : X \to Y$ be a morphism and let $t : Z \to X$ be an element of $S$ such that $a \circ t = 0$. To prove LMS3 it suffices to find an $s \in S$ such that $s \circ a = 0$, compare with the proof of Lemma \ref{lemma-triangle-functor-localize}. Choose a distinguished triangle $(Z, X, Q, t, g, h)$ using TR1 and TR2. Since $a \circ t = 0$ we see by Lemma \ref{lemma-representable-homological} there exists a morphism $i : Q \to Y$ such that $i \circ g = a$. Finally, using TR1 again we can choose a triangle $(Q, Y, W, i, s, k)$. Here is a picture $$ \xymatrix{ Z \ar[r]_t & X \ar[r]_g \ar[d]^1 & Q \ar[r] \ar[d]^i & Z[1] \\ & X \ar[r]_a & Y \ar[d]^s \\ & & W } $$ Since $t \in S$ we see that $Q$ is isomorphic to an object of $\mathcal{D}'$. Hence $s \in S$. Finally, $s \circ a = s \circ i \circ g = 0$ as $s \circ i = 0$ by Lemma \ref{lemma-composition-zero}. We conclude that LMS3 holds. The proof of RMS3 is dual. \medskip\noindent Proof of MS5. Follows as distinguished triangles and $\mathcal{D}'$ are stable under translations \medskip\noindent Proof of MS6. Suppose given a commutative diagram $$ \xymatrix{ X \ar[r] \ar[d]^s & Y \ar[d]^{s'} \\ X' \ar[r] & Y' } $$ with $s, s' \in S$. By Proposition \ref{proposition-9} we can extend this to a nine square diagram. As $s, s'$ are elements of $S$ we see that $X'', Y''$ are isomorphic to objects of $\mathcal{D}'$. Since $\mathcal{D}'$ is a full triangulated subcategory we see that $Z''$ is also isomorphic to an object of $\mathcal{D}'$. Whence the morphism $Z \to Z'$ is an element of $S$. This proves MS6. \medskip\noindent MS2 is a formal consequence of MS1, MS5, and MS6, see Lemma \ref{lemma-localization-conditions}. This finishes the proof of the first assertion of the lemma. \medskip\noindent Let's assume that $S$ is saturated. (In the following we will use rotation of distinguished triangles without further mention.) Let $X \oplus Y$ be an object isomorphic to an object of $\mathcal{D}'$. Consider the morphism $f : 0 \to X$. The composition $0 \to X \to X \oplus Y$ is an element of $S$ as $(0, X \oplus Y, X \oplus Y, 0, 1, 0)$ is a distinguished triangle. The composition $Y[-1] \to 0 \to X$ is an element of $S$ as $(X, X \oplus Y, Y, (1, 0), (0, 1), 0)$ is a distinguished triangle, see Lemma \ref{lemma-split}. Hence $0 \to X$ is an element of $S$ (as $S$ is saturated). Thus $X$ is isomorphic to an object of $\mathcal{D}'$ as desired. \medskip\noindent Finally, assume $\mathcal{D}'$ is a saturated triangulated subcategory. Let $$ W \xrightarrow{h} X \xrightarrow{g} Y \xrightarrow{f} Z $$ be composable morphisms of $\mathcal{D}$ such that $fg, gh \in S$. We will build up a picture of objects as in the diagram below. $$ \xymatrix{  & & Q_{12} \ar[rd] & & Q_{23} \ar[rd] \\  & Q_1 \ar[ld]_{\! + \! 1} \ar[ru] & & Q_2 \ar[ld]_{\! + \! 1} \ar[ll]_{\! + \! 1} \ar[ru] & & Q_3 \ar[ld]_{\! + \! 1} \ar[ll]_{\! + \! 1} \\ W \ar[rr] & & X \ar[lu] \ar[rr] & & Y \ar[lu] \ar[rr] & & Z \ar[lu] } $$ First choose distinguished triangles $(W, X, Q_1)$, $(X, Y, Q_2)$, $(Y, Z, Q_3)$ $(W, Y, Q_{12})$, and $(X, Z, Q_{23})$. Denote $s : Q_2 \to Q_1[1]$ the composition $Q_2 \to X[1] \to Q_1[1]$. Denote $t : Q_3 \to Q_2[1]$ the composition $Q_3 \to Y[1] \to Q_2[1]$. By TR4 applied to the composition $W \to X \to Y$ and the composition $X \to Y \to Z$ there exist distinguished triangles $(Q_1, Q_{12}, Q_2)$ and $(Q_2, Q_{23}, Q_3)$ which use the morphisms $s$ and $t$. The objects $Q_{12}$ and $Q_{23}$ are isomorphic to objects of $\mathcal{D}'$ as $W \to Y$ and $X \to Z$ are assumed in $S$. Hence also $s[1]t$ is an element of $S$ as $S$ is closed under compositions and shifts. Note that $s[1]t = 0$ as $Y[1] \to Q_2[1] \to X[2]$ is zero, see Lemma \ref{lemma-composition-zero}. Hence $Q_3[1] \oplus Q_1[2]$ is isomorphic to an object of $\mathcal{D}'$, see Lemma \ref{lemma-split}. By assumption on $\mathcal{D}'$ we conclude that $Q_3$ and $Q_1$ are isomorphic to objects of $\mathcal{D}'$. Looking at the distinguished triangle $(Q_1, Q_{12}, Q_2)$ we conclude that $Q_2$ is also isomorphic to an object of $\mathcal{D}'$. Looking at the distinguished triangle $(X, Y, Q_2)$ we finally conclude that $g \in S$. (It is also follows that $h, f \in S$, but we don't need this.)