[Math] Is there notation for “some two of the three statements are true”

logicnotationpropositional-calculus

There are three propositions A, B, C and another condition "some two of these propositions are true and the third one is false", or, in other words, "exactly 2 of 3 propositions are true". Using truth tables and a Karnaugh map (as discussed at How to find the logical formula for a given truth table?) i deducted the Boolean expression for this: ABC' + AB'C + A'BC. Is there any more succinct notation for this expression in any branch of logic?

Edit: Obviously using proposition calculus notation the above statement may be represented as: $(A \wedge B \wedge \neg C) \vee (A \wedge \neg B \wedge C) \vee (\neg A \wedge B \wedge C)$. I am sorry if that misguided you. I'm still interested, if any more succinct notation is possible.

Best Answer

Using the Iverson bracket, $$ [A]+[B]+[C]=2 $$

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[Math] Propositional Logic Help: $(\neg p \wedge (p \vee q)) \rightarrow q $ is a tautology

Hint: We have that $$ (\neg p \land q)\to q \equiv (p\lor \neg q)\lor q \equiv p \lor (\neg q\lor q). $$ You should be able to take it from there.

[Math] Boolean Simplification Methods. : Karnaugh Maps vs Boolean Algebra

Both Karnaugh Map and Boolean Algebra Simplification need not to give same answer. The answer may differ.

The Boolean Algebra Simplification is sometimes tricky because we need smart use of Properties (Absorption and Distributive) and Theorems (Redundancy Theorem)

Even K-Map Solutions are not unique. The answer may differ based on your choice of pairing. The same set of minterms can be simplified in two ways as shown in this attached image.

Now, considering your problem, all the three expressions are not equivalent.

Let three expressions be
$\mathcal{F=A′B′C′+A′B′C+A′BC′+ABC′+ABC}$
$\mathcal{K=A′B′C+AB}$
$\mathcal{S=A′+B}$

$\mathcal{A}$	$\mathcal{B}$	$\mathcal{C}$	$\mathcal{F}$	$\mathcal{K}$	$\mathcal{S}$
0	0	0	1	0	1
0	0	1	1	1	1
0	1	0	1	0	1
0	1	1	0	0	1
1	0	0	0	0	0
1	0	1	0	0	0
1	1	0	1	1	1
1	1	1	1	1	1

Clearly, $\mathcal{F}$, $\mathcal{K}$ and $\mathcal{S}$ aren't equivalent.

Thus, the claimed simplified expression (using K-Map and Boolean) are actually incorrect.

The actual simplified expressions are :

$\mathcal{AB+A′B′+A′C′}$

$\mathcal{AB+A′B′+BC′}$

Best Answer

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[Math] Propositional Logic Help: $(\neg p \wedge (p \vee q)) \rightarrow q $ is a tautology

[Math] Boolean Simplification Methods. : Karnaugh Maps vs Boolean Algebra

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