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The algebraic normal form is the unique multilinear polynomial representation of a Boolean function over the field with two elements, using exclusive or for addition and ordinary bit multiplication for products. The empty product is one. Existence and uniqueness follow by induction from $f(z^{\prime },z_d)=f(z^{\prime },0)\oplus z_d(f(z^{\prime },0)\oplus f(z^{\prime },1))$: the first polynomial supplies all coefficients not containing $z_d$, and the difference supplies exactly those that do. For split inputs $(x,y)$, each monomial factors into one locally computed product of Alice's bits and one of Bob's bits, yielding a separated Boolean decomposition.

For a stationary source of independent and identically distributed fair bits, with Hamming distortion and binary reconstruction alphabet,The binary entropy convention is $0{\log }_{2}0=0$. Once $D\ge 1/2$, always reconstructing zero meets the distortion constraint without sending any bits. The formula $1-h_2(D)$ must not be extended to $D>1/2$: a looser constraint cannot increase the required rate.^[ref]

The short argument identifies both sides of the optimum. For any feasible binary reconstruction, put $E=X\oplus \hat{X}$, using exclusive or. Knowing $\hat{X}$ makes $X$ and $E$ determine one another, so the conditional entropy obeyswhen $0\le D\le 1/2$. The last step uses the increase of binary entropy on that interval. Hence $I(X;\hat{X})=1-H(X\mid \hat{X})\ge 1-h_2(D)$. Conversely, choose $\hat{X}=X\oplus N$ with an independent Bernoulli random variable $N$ of parameter $D$. This binary symmetric channel has expected distortion $D$ and mutual information $1-h_2(D)$, attaining the bound.

A binary symmetric channel is a discrete memoryless channel that flips each transmitted bit with fixed crossover probability $p$. For one use, write $Y=X\oplus Z$, where $\oplus$ is exclusive or and $Z$ is a Bernoulli random variable with parameter $p$, independent of the input $X$. Thus $Z=0$ preserves the bit and $Z=1$ flips it. Different uses have independent noise under this model.

With row labels $X=0,1$ and column labels $Y=0,1$, its transition matrix isEach row sums to one, and exchanging both binary labels leaves the rule unchanged, explaining symmetric. For $p=0.1$, a prescribed three-bit input $010$ produces $010$ with probability $0.9^3=0.729$ and $011$ with probability $0.9^2(0.1)=0.081$. The received $011$ does not itself announce the changed position: it could also arise from input $011$ with no flips.

The binary symmetric channel capacity quantifies how well a block code can overcome this uncertainty, and input distribution optimization for a binary symmetric channel shows why a fair input attains the best information rate. A completely noisy binary symmetric channel has $p=1/2$ and carries no input information; a known flip on every use, $p=1$, is instead reversible.

A Boolean operation is a Boolean function taking truth values encoded as bits: $0$ means false and $1$ means true. These operations supply the concrete two-valued interpretation of a Boolean algebra. A Boolean valuation fixes the inputs; a truth table lists every possible input. With two input bits there are four rows.

The basic logical connectives are logical conjunction (AND, $P\wedge Q$, both true), logical disjunction (OR, $P\vee Q$, at least one true), exclusive or (XOR, $P\oplus Q$, exactly one true), and logical negation (NOT, $\neg P$, the opposite value). OR is inclusive: it accepts the case where both inputs are true.^[ref]

For an original program-condition example, let $P$ mean “the backup is recent” and $Q$ mean “the backup is verified”. A rule requiring a recent, verified backup uses logical conjunction; one accepting either qualification uses logical disjunction. A rule flagging a mismatch between the two qualifications uses exclusive or. A recent, verified backup gives $(P,Q)=(1,1)$: AND and OR accept it, while XOR does not flag it. The last row of the truth table makes this difference explicit.

The same small Boolean operations build larger tests and Boolean circuits. In a programming language, the value of a condition and the procedure for evaluating it both matter. For Boolean operands, Python's and, or, and not implement these logical connectives. Evaluation proceeds from the left; and skips its right operand if the left is false, and or skips it if the left is true. More generally, and and or return an operand, whereas not returns a Boolean.^[ref]

For example, the Python condition denominator != 0 and numerator / denominator > 2 avoids division by zero because the logical conjunction's second test is evaluated only when the first succeeds. Reversing the tests preserves the abstract truth table for defined operands but can raise a division-by-zero error handled through exception handling. A rewrite involving a side effect must also preserve the needed evaluation behavior.

Events turn Boolean operations into operations on sets. For a die with sample space $U=\{1,2,3,4,5,6\}$, let $A=\{2,4,6\}$ be the event “even outcome” and $B=\{4,5,6\}$ the event “outcome at least four”. At outcome $u$, use the indicator functions $P=1_A(u)$ and $Q=1_B(u)$. ThenThus logical conjunction becomes set intersection, logical disjunction becomes set union, exclusive or becomes symmetric difference, and logical negation becomes the complement of a set within $U$. Outcome four satisfies both events, so inclusive logical disjunction includes it and exclusive or excludes it. If the die is fair, these four events have probabilities $1/3$, $2/3$, $1/3$, and $1/2$, respectively. No assumption of independent events is needed for the set identities; indeed $A$ and $B$ here are not independent, since $1/3\ne (1/2)(1/2)$.

For finite discrete random variables, the mutual information supplied by two observations equals the information supplied by the first plus the additional conditional mutual information supplied by the second:The order changes the individual contributions but leaves the total unchanged.^[ref] To prove the first identity, add and subtract the intermediate conditional entropy:With base-two logarithms, all contributions are in bits. Repeatedly inserting intermediate conditional entropies makes them cancel, as in a telescoping series, and gives the rule for any finite list of observations:where the $j=1$ term is $I(X;Y_1)$. The same calculation with $W$ already known gives $I(X;Y,Z\mid W)=I(X;Y\mid W)+I(X;Z\mid Y,W)$.

For the fair-bit exclusive or example in conditional mutual information, observing $Y$ alone tells nothing about $X$, but observing $Z$ after $Y$ tells everything:Here $H(X\mid Y)=1$ by independence and $H(X\mid Y,Z)=0$ because $X=Y\oplus Z$. In the reverse order, $H(X\mid Z)=1$, so $I(X;Z)=0$ and $I(X;Y\mid Z)=1$: the two observations are individually uninformative but jointly reveal one bit. Simply adding their separate mutual informations would miss all of it.

The chain rule is useful for measuring the value of each new observation without double counting information already supplied. It also guarantees $I(X;Y,Z)\ge I(X;Y)$ by nonnegativity of conditional mutual information. Keeping $Y$ and adding $Z$ can help, even though discarding $Y$ and retaining only a processed $Z$ is constrained by the data processing inequality. Equating the two observation orders gives the precise identitywhich explains why there is no general rule that conditioning decreases mutual information.

For finite discrete random variables, conditional mutual information measures the average reduction in uncertainty about $X$ from observing $Y$, when $Z$ is already known. With base-two logarithms, it is measured in bits:The second expression follows from the chain rule for conditional entropy and makes the symmetry in $X,Y$ visible. Equivalently, compute the mutual information separately in each conditional probability distribution and then average:Here $D$ is the Kullback-Leibler divergence, also computed with base-two logarithms; zero-probability terms contribute zero. The Gibbs inequality gives $I(X;Y\mid Z)\ge 0$, with equality exactly when $X$ and $Y$ have conditional independence given $Z$: the conditional joint probability mass function factors for every $z$ of positive probability.^[ref]

For a fully computed example, take independent random variables $X,Y$ that are fair bits, and let $Z=X\oplus Y$, where $\oplus$ is exclusive or. The complete joint probability distribution has only these four positive entries:

Every pair $(x,y)$ has probability $1/4=p_X(x)p_Y(y)$, so $I(X;Y)=4(1/4){\log }_{2}1=0$. But $p_Z(0)=p_Z(1)=1/2$. Given $Z=0$, the pairs $(0,0),(1,1)$ each have conditional probability $1/2$, while both conditional marginals are fair. ThusGiven $Z=1$, the pairs $(0,1),(1,0)$ give the same calculation. ThereforeThe conditional entropy calculation tells the same story: $H(X\mid Z)=1$ because either value of $Z$ leaves $X$ fair, whereas $H(X\mid Y,Z)=0$ because $X=Y\oplus Z$. Learning the exclusive or does not reveal either bit alone; it reveals whether they agree, so learning one then determines the other.

This example distinguishes conditional independence from ordinary independence. It also disproves the claim that conditioning always reduces mutual information, despite conditioning reduces entropy. There is no general ordering between $I(X;Y\mid Z)$ and $I(X;Y)$: the example above increases it from zero to one, while taking $Y=X$ and $Z=X$ for a fair bit decreases it from one to zero. The chain rule for mutual information explains why both are possible.^[ref]

Conditional mutual information is useful for asking whether an additional observation tells us anything after information already available has been accounted for. Its zero case expresses conditional independence exactly; in a Markov chain, this is the step that turns the chain rule for mutual information into the data processing inequality.

Return to pairwise independence does not imply joint independence, where $X,Y$ are independent fair bits and $Z=X\oplus Y$ is their exclusive or. Given $Z=0$, only $000$ and $110$ remain, each with conditional probability $1/2$. Given $Z=1$, only $011$ and $101$ remain, again each with conditional probability $1/2$. Consequently,The conditional marginal distributions are still fair, so their product has four entries $1/4$ in either case. The actual tables differ: $Y=X$ given $Z=0$, and $Y=1-X$ given $Z=1$. Thus unconditional independence need not survive conditioning. Observing whether two bits agree restricts their possible combinations; no physical interaction between the original trials is required to produce that information constraint.

Let $A$ be a fair bit, let $N$ be an independent Bernoulli random variable with $\mathbb{P}(N=1)=p$, and set $B=A\oplus N$, where $\oplus$ is exclusive or. Thus $N=1$ flips the first bit. This constructs a binary symmetric channel with a fair input, for $0\le p\le 1$. The joint probability mass function isThe column sums show that $B$ is also a fair bit, whatever $p$ is. Both marginal distributions therefore have information entropy one bit. Given $A$, only the flip decision remains unknown, giving conditional entropy $H(B\mid A)=h_2(p)$, whereis the binary entropy, with $h_2(0)=h_2(1)=0$. The chain rule for information entropy and the definition of mutual information now give

Take $p=1/10$. The four joint probabilities, in row order, are $(0.45,0.05,0.05,0.45)$. Directly from the definition of joint entropy,Conditioning on either value of $A$ gives probabilities $0.9$ and $0.1$ for the two values of $B$, henceAdding $H(A)+H(B)=2$ would overcount by exactly $I(A;B)$. These are average uncertainties under the stipulated probability distribution; the four possible pairs are still represented by four distinct outcomes.

There is another way to see the decomposition: $(A,B)\mapsto (A,A\oplus B)=(A,N)$ is a bijection. It merely relabels the four outcomes, so it preserves joint entropy. The independence of $A$ and $N$ then gives $H(A,B)=H(A,N)=H(A)+H(N)=1+h_2(p)$. Describing the initial bit and the flip decision separates shared uncertainty from new uncertainty.

At $p=0$, the bits are identical: $B=A$, so $H(B\mid A)=0$ and $I(A;B)=1$. At $p=1$, they are complementary: $B=1-A$, with exactly the same conditional entropy and mutual information. In either case, observing one determines the other. At $p=1/2$, all four joint probabilities are $1/4$; the bits are independent, $H(A,B)=2$, and $I(A;B)=0$. Perfect disagreement is therefore just as informative as perfect agreement.

For $p>1/2$, guessing $A=1-B$ succeeds with probability $p$, better than guessing $A=B$. Changing $B$ to $1-B$ replaces $p$ by $1-p$ and preserves mutual information, since the relabeling is invertible. What matters is how well the observation distinguishes the input, rather than how often its symbol matches the input. The same symmetry makes the binary entropy curve peak at independence, while the mutual information curve reaches its minimum there.

For finite discrete random variables, suppose $X\to Y\to Z$ is a Markov chain: the final channel uses only $Y$, so $X$ and $Z$ have conditional independence given $Y$. In terms of the joint probability mass function,and hence $I(X;Z\mid Y)=0$. Then the two data processing inequalities areAll mutual informations here use base-two logarithms. Thus deterministic or randomized processing cannot increase the information shared with an unprocessed variable.^[ref]

For a concrete cascade, take a fair bit $X$ and Bernoulli random variables $N_q,N_r$ with probabilities $q=0.1$, $r=0.2$ of being one. Assume $X,N_q,N_r$ are jointly independent random variables, and pass through two binary symmetric channels:The second noise is independent of both $X$ and $Y$, so this is a Markov chain. Both $Y$ and $Z$ are fair bits. The net exclusive or error is one exactly when one noise flips the bit and the other does not:Using the binary entropy $h_2(t)=-t{\log }_{2}t-(1-t){\log }_2(1-t)$, the three mutual informations are fully determined:All numbers are in bits. The two losses are $I(X;Y\mid Z)\approx 0.3577507789$ and $I(Y;Z\mid X)\approx 0.1048182776$. Thus the cascade retains less information about the original source than either individual channel shares between its own input and output.

With $q=0.1$ fixed and $0\le r\le 1/2$, the effective error is $s=0.1+0.8r$. At $r=0$, $Z=Y$ and $I(X;Z)=I(X;Y)$; at $r=1/2$, $Z$ is independent of the previous variables and $I(X;Z)=I(Y;Z)=0$. The diagram plots these endpoints and the computed $r=0.2$ example.

The data processing inequality is useful for testing what a proposed decoder or data summary could possibly reveal: if its only input is $Y$, even randomized processing cannot provide more mutual information about $X$ than $Y$ already contains. The Markov chain hypothesis matters. In the exclusive or example from conditional mutual information, allowing a processor to use both $Y$ and $Z$ reveals $X$ exactly; that processor has additional input, so the conclusion for processing $Y$ alone does not apply.

The chain rule for mutual information proves both inequalities and identifies their exact gaps. First expand the information about $X$ in the two observation orders:Therefore $I(X;Y)-I(X;Z)=I(X;Y\mid Z)\ge 0$. Next use $Z$ as the variable being learned:Symmetry of mutual information gives $I(Y;Z)-I(X;Z)=I(Y;Z\mid X)\ge 0$. Both signs follow from the nonnegativity of conditional mutual information; the first gap also gives equality in data processing inequality.

Exclusive or returns one exactly when its two input bits differ. Its truth table gives $0\oplus 0=0$, $0\oplus 1=1$, $1\oplus 0=1$, and $1\oplus 1=0$. The identitiesexpress exclusive or using logical conjunction, logical disjunction, and logical negation. The first expression accepts at least one input and then excludes both; the second explicitly lists the two accepted input pairs.

For numerical bits, exclusive or is addition in modular arithmetic modulo two:The binary operation is commutative and associative, because addition modulo two has those properties. It satisfies $x\oplus 0=x$, $x\oplus 1=\neg x$, and $x\oplus x=0$. These identities explain why a bit can be toggled and restored with the same operation.

On binary strings of equal length, bitwise exclusive or applies the one-bit rule separately at each position. For example,A bit mask of $0110$ toggles exactly the middle two bits. Applying the mask twice restores the original binary string: each selected bit changes twice, and each unselected bit stays fixed. This positional operation is distinct from taking the exclusive or of the overall truth values of two integers.