Proof.
Let X_1 \to X_2 \to X_3 \to X_1[1] be a distinguished triangle and assume P holds for X_1 and X_2. Choose distinguished triangles
A_1 \to X_1 \to B_1 \to A_1[1] \quad \text{and}\quad A_2 \to X_2 \to B_2 \to A_2[1]
as in condition P. Since \mathop{\mathrm{Hom}}\nolimits (A_1, A_2) = \mathop{\mathrm{Hom}}\nolimits (A_1, X_2) by Lemma 13.40.2 there is a unique morphism A_1 \to A_2 such that the diagram
\xymatrix{ A_1 \ar[d] \ar[r] & X_1 \ar[d] \\ A_2 \ar[r] & X_2 }
commutes. Choose an extension of this to a diagram
\xymatrix{ A_1 \ar[r] \ar[d] & X_1 \ar[r] \ar[d] & Q_1 \ar[r] \ar[d] & A_1[1] \ar[d] \\ A_2 \ar[r] \ar[d] & X_2 \ar[r] \ar[d] & Q_2 \ar[r] \ar[d] & A_2[1] \ar[d] \\ A_3 \ar[r] \ar[d] & X_3 \ar[r] \ar[d] & Q_3 \ar[r] \ar[d] & A_3[1] \ar[d] \\ A_1[1] \ar[r] & X_1[1] \ar[r] & Q_1[1] \ar[r] & A_1[2] }
as in Proposition 13.4.23. By TR3 we see that Q_1 \cong B_1 and Q_2 \cong B_2 and hence Q_1, Q_2 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}^\perp ). As Q_1 \to Q_2 \to Q_3 \to Q_1[1] is a distinguished triangle we see that Q_3 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}^\perp ) by Lemma 13.40.4. Since \mathcal{A} is a full triangulated subcategory, we see that A_3 is isomorphic to an object of \mathcal{A}. Thus X_3 satisfies P. The other cases of (1) follow from this case by translation. Part (2) is a special case of (1) via Lemma 13.4.11.
\square
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