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Assume that a language A is reducible to language B. The claim is true?

if A is decidable then B is decidable too.

The correct answer is:

This claim is wrong. If A is e.g. the empty language (which is clearly decidable) and B is $A_{TM}$, then surely ∅ is reducible to $A_{TM}$, but $A_{TM}$ is undecidable. The claim is true the other way round: If B is decidable then A is decidable too.

I have problem to understand two points:

  1. A is e.g. the empty language (which is clearly decidable)

Why empty language is ("clearly") decidable and $L_∅$ is not?, where, $$L_\emptyset = \{\langle M\rangle \mid M \text{ is a Turing Machine and }L(M)=\emptyset\}.$$

  1. **A is reducible to $A_{TM}$ **

I have difficulties to image a reduction from $∅$ to $A_{TM}$, where $A_{TM} = \{<M,w> | \ M \ is \ a \ TM \ and \ M \ accepts \ w\}$

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  • $\begingroup$ In which relation are A and B? What is ATM? $\endgroup$ – Daniel Dec 26 '19 at 17:34
  • $\begingroup$ language A is reducible to language B. Added on the question. Thanks. $\endgroup$ – Gianni Spear Dec 26 '19 at 17:37
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  1. By definition, a language $L$ is decidable if there exists a TM $M\mid L = L(M)$ deciding it. Consider a TM that rejects on all inputs (for example, one where $q_o = q_{\text{rej}}$). The language of this TM is $\emptyset$, so $L=\emptyset$ is decidable.
  2. With reductions, in general if $A\leq B$ ($A$ reduces to $B$), then $B$ is at least as hard as $A$ with respect to decidability. An undecidable language $A_{TM}$, regardless of what it actually is, is harder to decide than the decidable language $\emptyset$ by definition. The specifics of the reduction don't matter; the only information we needed was given in the problem statement (that $A$ is reducible to $B$).
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    $\begingroup$ I don't think the empty language is a good example for the answer because there is not reduction from the empty language to an another language. $\endgroup$ – Gianni Spear Dec 26 '19 at 18:17
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    $\begingroup$ By Turing reductions specifically, the empty language can be reduced to other languages. Let $M_B$ be the decider for $B$; then the reduction from $A = \emptyset$ to $B$ could just be "def $M_A\langle x\rangle$: run $M_B\langle x\rangle$, then reject." $\endgroup$ – elucidium Dec 26 '19 at 18:33
  • $\begingroup$ @GianniSpear That's not correct (and this answer is right). As elucidium says, it's easy to whip up a Turing reduction from $\emptyset$ to $A$ by ignoring the oracle $A$ completely. We can build even stronger reductions too. For example, we have $\emptyset\le_mA$ for any $A$ (where $\le_m$ refers to many-one reducibility): if $A=\emptyset$ we just use the identity function, and if $A\not=\emptyset$ we pick some $n\in A$ and use the constant function $x\mapsto A$. It's only when we consider $1$-reductions that things break down. $\endgroup$ – Noah Schweber Dec 26 '19 at 19:31
  • $\begingroup$ (FWIW it's still the case that $\emptyset\le_1A$ whenever the complement of $A$ has an infinite computable subset - and $A_{TM}$ fits this criterion via the Padding Lemma. So in fact $\emptyset\le_1A_{TM}$, which is about as strong a fact as you could hope for.) $\endgroup$ – Noah Schweber Dec 26 '19 at 19:34
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Elucidium addresses most of your question; let me address the remaining point, which is why $\emptyset$ and $L_\emptyset$ behave differently.

The point is that they're simply very different sets in the first place. For example, $\emptyset$ has no elements - that's its definition - while $L_\emptyset$ is infinite (there are lots of Turing machines which don't do anything). Just because something is related to $\emptyset$ doesn't mean it behaves like $\emptyset$.

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