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$A \land B \to C$ is true. $C \to A$ is true. Is it always possible to find such $D$ that $B \to D$ is true and $A \land D \leftrightarrow C$ is true?

To clarify my question I give an example. If a space is regular and second countable then it is metrizable. If a space is metrizable then it is regular. A space is metrizable if and only if it is regular and has locally finite countable basis.

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    why use such an example and not just try a truthtable? – 2017-01-17
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    One issue that can be confusing is whether you want to view $A,B,C$ as statements with a particular truth value (e.g. looking at a particular space) or whether you are actually looking at $(\forall x)[A(x) \land B(x) \to C(x)]$, $(\forall x)[C(x) \to A(x)]$, etc. In the first sentence, there are no quantifiers, but in the second sentence there are implicit universal quantifiers. These have a significant impact on the underlying logical structure. – 2017-01-17
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    $$D:=C{}{}{}{}$$ – 2017-01-17
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    There are only 256 boolean functions of $A$, $B$, and $C$. You could just check all of them. – 2017-01-17
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    @Did: that does work, but in general the weakest solution modulo $A$ is $D := B \lor C$. – 2017-01-17
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    @CarlMummert Of course. Note that the OP asks for some set $D$, any of them. – 2017-01-17

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Finding $D$ such that $B \rightarrow D$ and $A \wedge D \rightarrow C$ if $A \wedge B \rightarrow C$ and $C \rightarrow A$ means solving a Boolean equation. Some people call this problem Boolean unification; others call it synthesis of (Boolean) Skolem functions. The algorithm is due to Boole himself. In this case the equation is

$$ ((A \wedge B \rightarrow C) \wedge (C \rightarrow A)) \rightarrow ((B \rightarrow D) \wedge (A \wedge D \leftrightarrow C)) = \top\enspace.$$

This equation is (unconditionally) consistent, which means that there are functions of $A$, $B$, and $C$ that, when substituted for $D$, make the left-hand side tautologous. The most general solution (or unifier) in parametric form is:

$$D = (A \wedge C) \vee (\neg A \wedge B \wedge \neg C) \vee (p \wedge (\neg A \vee B)) \enspace, $$

where $p$ is a parameter that ranges over the functions of $A$, $B$, and $C$. Particular solutions may be obtained by choosing values of $p$. For instance, for $p = B \vee C$, one obtains $D = B \vee C$. This is no coincidence: if $\tilde{D}$ is a particular solution, it is obtained by setting $p=\tilde{D}$ (and perhaps by other values of $p$ too).

From the most general solution, the strongest and weakest solutions are obtained for $p=\bot$ and $p=\top$, respectively. (The solution is always unate positive in $p$.)

$$\begin{align} D_{\bot} &= (A \wedge C) \vee (\neg A \wedge B \wedge \neg C) \\ D_{\top} &= \neg A \vee B \vee C \enspace. \end{align}$$

In the simple case of one unknown ($D$ for us) the consistency is $\exists D \,.F = F_{D} \vee F_{\neg D}$, where $F$ is the left-hand side of the equation, $F_D$ is $F$ with $D$ set to true, and $F_{\neg D}$ is $F$ with $D$ set to false. If the quantification results in a tautology---as in our case---the equation is (unconditionally) consistent. The parametric solution is then $$ D = \neg F_{\neg D} \vee (p \wedge F_D) \enspace.$$

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Here is a more principled solution. We have:

  • $C \to A$ : so $C$ is "stronger" than $A$
  • $A \land B \to C$: so there is something which joins with $A$ to form a statement stronger than $C$. The problem is that $A \land B$ might be strictly stronger than $C$.

In this case, what we need to do is to weaken $A \land B$ to something equivalent to $C$. Normally, we would weaken a statement with a $\lor$. In particular, we have $(A \land B) \lor C \leftrightarrow C$. However, $(A \land B) \lor C$ is not of the form $A \land D$.

We can use the distributive law to convert $(A \land B)\lor C$ to $(A \lor C) \land (B \lor C)$. That is closer to what we want.

Now $A \lor C$ implies $A$, because $C$ implies $A$, and $A$ implies $A \lor C$, so we have $(A \lor C) \leftrightarrow A$.

Therefore, we have $$ A \land (B \lor C) \leftrightarrow (A \lor C) \land (B \lor C) \leftrightarrow (A \land B) \lor C \leftrightarrow C. $$

So we can take $D$ to be $B \lor C$. In general, $B \lor C$ is no stronger than $C$, but it could be strictly weaker.

Going back to the example in the question, "If a space is regular and second countable then it is metrizable. If a space is metrizable then it is regular.", we have

  • A = "regular"
  • B = "second countable"
  • C = "metrizable"
  • D = "metrizable or second countable", which is strictly weaker than $C$

so we have $C \leftrightarrow A \lor D$: "a space is metrizable if and only if it is regular and (metrizable or second countable)". Unfortunately, this is not very informative, which is expected. To get more informative equivalences, you'd have to look at what the words actually mean, rather than merely pushing around the logical connectives.

Finally, we can ask whether $B \lor C$ is the "weakest" solution the question. It is indeed, modulo $A$. In other words, if we have $A \land E \leftrightarrow C$ then $A \to (E \leftrightarrow B \lor C)$. For one direction, because $A \land E \to C$, we immediately have $A \to (E \to B \lor C)$. For the other direction, because $A \land (B \lor C)$ implies $C$, and $C$ implies $E$, so $A \to ( B \lor C \to E)$.

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    You have simplified my question. $D \not \to B$ is true of course. I have just added this extra condition to my question. – 2017-01-17
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    @alch: I don't follow; $D = B \lor C$ does not imply $B$ in general. For example "metrizable or second countable" does not imply "second countable". Similarly it does not imply $C$ in general. – 2017-01-17
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    You are right. I have changed the extra condition to this $B \to D$ is true. – 2017-01-18