Every oracle is strictly below its jump

Let $Q_n$ be the oracle program that queries $A$ at $n$ and halts exactly when the answer is yes. Inserting the numeral $n$ into this fixed program template gives, by the hardwiring compiler in The Turing jump, a total computable index map $q(n)$. Hence $n\in A⟺{\mathrm{\Phi }}_{q(n)}^A(q(n))\downarrow ⟺q(n)\in A^{\mathrm{\prime }},$ so $A{\le }_T{A}^{\mathrm{\prime }}$.

Suppose toward contradiction that an $A$-oracle program $D$ decides $A^{\mathrm{\prime }}$. Build an oracle program $Q$ that, on input $z$, runs $D^A(z)$ and halts exactly when $D$ rejects. Let $e$ be the fixed index of $Q$ in the enumeration of The Turing jump. Since $D$ is correct, $e\in A^{\mathrm{\prime }}⟺{\mathrm{\Phi }}_e^A(e)\downarrow ⟺D^A(e)=0⟺e\mathrm{\notin }{A}^{\mathrm{\prime }},$ a contradiction.

Thus $A^{\mathrm{\prime }}\mathrm{̸}{\le }_{T}A$, while step 1.1 gives $A{\le }_T{A}^{\mathrm{\prime }}$; therefore $A{<}_T{A}^{\mathrm{\prime }}$.