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A memory operation's execution is conditional on a preceding value. A control dependency alone does not generally order two reads on POWER architecture or the relevant ARM memory models; it must not be substituted for an address dependency in pointer-publication reasoning.

For a measure space $(\Omega ,\mathcal{F},\mu )$ and $E\in \mathcal{F}$, the restriction of a measure to a measurable subset is the measure on $(E,\mathcal{F}{|}_E)$ given by the displayed formula. The trace sigma-algebra consists of the measurable sets contained in $E$. The empty-set axiom and countable additivity are inherited directly from $\mu$. An associated measure on the original sigma-algebra is $\nu (A)=\mu (A\cap E)$ for $A\in \mathcal{F}$; the two constructions have different underlying measurable spaces.

For measurable sets $A\subseteq B$, the disjoint union $B=A\bigsqcup (B\setminus A)$ and finite additivity give $\mu (B)=\mu (A)+\mu (B\setminus A)\ge \mu (A)$. This uses nonnegativity and is valid with infinite values; it does not subtract infinities. Monotonicity of a measure implies that increasing measurable sets have nondecreasing measures, but the equality of their limit with the measure of their union uses the stronger axiom of countable additivity.

A Lebesgue measurable function $f:\mathbb{R}\to \mathbb{R}$ has a Lebesgue measurable set as the preimage of every open subset of $\mathbb{R}$. Equivalently, use every Borel set in the target: the sets whose preimages are measurable form a sigma-algebra containing all open sets. The source carries the Lebesgue sigma-algebra and the target its Borel sigma-algebra. A Borel measurable function on $\mathbb{R}$ instead requires the preimages to be Borel in the source as well. Restricting the source to a Lebesgue measurable set uses its induced sigma-algebra.

If $u,v$ are harmonic functions on a bounded domain and continuous on its closure, their difference is a harmonic function. Its minimum is attained on the boundary by the maximum principle for harmonic functions applied to its negative. Nonnegative boundary difference therefore implies nonnegative interior difference. Applying the comparison in both directions proves uniqueness for the Dirichlet problem whenever a classical solution exists. This is an order statement and does not itself establish existence.

Root a finite tree. The two dynamic programming states are the optimal size in the descendant subtree when its root $v$ is included, $I(v)$, or excluded, $O(v)$. The sums run over its children. Inclusion forces every child to be excluded; exclusion permits each child's better state. Child subtrees have no edges between them, which proves independence of these choices. The leaf states are $I=1,O=0$, and the root's better state gives the optimum.

A postorder traversal computes the states in $O(n)$ time and storage. Reconstruct a set by following an attaining root choice and the same child constraints, breaking ties arbitrarily. The empty and singleton cases use the ordinary non-strict subset convention.

A propositionally consistent theory is maximally consistent when every proper extension by a new propositional formula is inconsistent. It decides each propositional formula or its negation.

The symmetry group of a hyperelastic material consists of invertible reference transformations $H$ preserving its strain energy density: $W(FH)=W(F)$ for every admissible deformation gradient $F$. A convention restricting reference transformations to determinant one restricts $H$ to the special linear group. Material isotropy requires all proper rotations in this group. This right action differs from the left action in material frame indifference.

For a smooth objective isotropic hyperelastic material, the strict Baker-Ericksen inequalities require larger distinct principal stretches to have larger principal stresses. For $W=\Phi (v_1,v_2,v_3)$ and $J=v_1{v}_2{v}_3$, ${\sigma }_i=J^{-1}{v}_i{\Phi }_i$. The strict convention implies every stress-free state has equal stretches: if $v_i\ne v_j$ but both stresses are zero, the displayed product would vanish. Thus a natural state of a hyperelastic material has $F=vQ$, $v>0$, $Q\in SO(3)$. A weak non-strict inequality alone does not imply this rigidity.

Let $Z$ be a nonnegative integer-valued random variable with $\mathbb{E}Z=A/q$, where $A\ge 0$ and $q\ge 1$ are integers. On $Z>A/q$, the integer $qZ$ is at least $A+1$. Markov's inequality gives $\mathrm{Pr}(Z>A/q)\le A/(A+1)$. Independent restarts therefore need at most $A+1$ attempts in expected value to achieve $Z\le \mathbb{E}Z$. The denominator bound on the expected value matters for this runtime guarantee.

A model of random exchange between two containers, associated with Paul Ehrenfest and Tatyana Afanasyeva-Ehrenfest. In the discrete-time Ehrenfest urn model, choose one of $N$ balls uniformly at each step and change its container. The count in the first container decreases with probability $r/N$ and increases with probability $(N-r)/N$. A continuous-time version lets each particle change container independently at rate $\lambda$, giving birth-death process transition rates $r\lambda$ and $(N-r)\lambda$. Its sequence of visited states is the discrete urn Markov chain. Stopping when one container becomes empty produces an absorbing variant, rather than changing the transition rule of the unstopped model.

An irreducible Markov chain is transient when its states are transient states. Starting at any one state, there is a positive probability of never returning, and the number of visits to any fixed state is finite almost surely.

For a time-homogeneous Markov chain started at $i$, condition on a return to $i$ at a fixed later time $N$, assuming that event has positive probability. The Markov property factors the probability of a visit to $j$ at time $n$ and a later return into the two transition probabilities in the numerator. Conditional probability then gives the formula. At the midpoint with $j=i$ and $N=2n$, it reduces to $p_n^2/p_{2n}$, where $p_n=(P^n{)}_{ii}$. If $p_n\sim a{\lambda }^n$ for $a,\lambda >0$, this midpoint return probability tends to $a$ even when unconditional returns tend to zero.

Let ${\xi }_1,{\xi }_2,\dots$ be independent and identically distributed random variables, independent of an initial state $Z_0$, and let $h$ be a measurable function. The displayed recursion defines a time-homogeneous Markov chain: conditioning on the state history leaves ${\xi }_{n+1}$ with its original probability distribution, so its next-state law depends only on $Z_n$. Its Markov kernel is $K(z,A)=\mathbb{P}(h(z,{\xi }_1)\in A)$. Running sums use $h(z,x)=z+x$; running maxima use $h(z,x)=\max (z,x)$. Independence from the complete past, rather than merely from the current state, supplies the Markov property.

A finite stock decreases by choosing a uniformly random positive number of its remaining items at every step. The remaining stock is a Markov chain with $P(i,j)=1/i$ for $0\le j<i$ and $P(0,0)=1$. State $0$ is an absorbing state, and every positive state is a transient state. If $e_i$ denotes the expected number of positive-stock steps before absorption, a first-step decomposition gives $e_0=0$ and $i{e}_i=i+{\sum }_{j<i}{e}_j$. Subtracting consecutive equations yields $e_i-e_{i-1}=1/i$, so $e_i=H_i$, the $i$th harmonic number.

The empirical occupation proportion is the fraction of the observed Markov chain states equal to $j$. These nonnegative fractions sum to one. To simulate them from a transition matrix, sample an initial state and then draw each successive state from the current state's row, accumulating visit counts. They are empirical frequencies for a finite path, whether or not the chain is already stationary. Long-run convergence to a stationary distribution requires the corresponding recurrence and irreducibility hypotheses.

Let $(X_n)$ be a time-homogeneous Markov chain with transition matrix $P$, and let $d$ be a positive integer. Then $Y_n=X_{dn}$ is a Markov chain with transition matrix $P^d$ and the same initial distribution. The Markov property and the Chapman-Kolmogorov equation give $P(X_{d(n+1)}=j\mid X_0,\dots ,X_{dn})=(P^d{)}_{X_{dn},j}$. Conditioning again on the smaller history $(Y_0,\dots ,Y_n)$ leaves the right-hand side unchanged, since it is determined by $Y_n$. This proves the Markov property of the sampled sequence, rather than only identifying its marginal transition probabilities. Periodicity of the original chain may split the sampled chain into several closed classes.

On the finite path $\{0,\dots ,m\}$, take interior neighbor probabilities $p$, endpoint inward probability $2p$, and holding probability $1-2p$, with $0<p\le 1/2$. Detailed balance gives the displayed invariant distribution: endpoint weights are half the interior weights. For $p<1/2$ the chain is irreducible and aperiodic, so transition probabilities converge to it. At $p=1/2$ parity alternates and the period is two.

From state zero the chain always goes to one. From a positive integer $j$, it resets to zero with probability $p$ and goes to $j+1$ otherwise. The stationary equations give ${\pi }_1={\pi }_0$ and ${\pi }_j=(1-p{)}^{j-1}{\pi }_0$, and normalization gives the displayed distribution. The chain is irreducible and positive recurrent; its return lengths two and three also make it aperiodic. The long-run proportion of time above a positive integer $K$ is $(1-p{)}^K/(1+p)$.

A two-phase inspection model alternates between inspecting every item and inspecting each item with probability $1/r$, where $r>1$. Suppose item defects are independent with probability $p\in (0,1)$. The first phase ends after $K\ge 1$ good items have been accumulated, while the second ends upon detecting a defect. With state $i<K$ equal to the cumulative number of good items found during the current full-inspection phase, the Markov chain has transitions $P_{ii}=p$, $P_{i,i+1}=q=1-p$, $P_{K0}=p/r$, $P_{KK}=1-p/r$.

Its stationary distribution is ${\pi }_i=p/(Kp+rq)$ for $i<K$ and ${\pi }_K=rq/(Kp+rq)$. Indeed, balance gives equal mass to every phase-A state and $p{\pi }_K/r=q{\pi }_{K-1}$; normalization fixes the common mass. The finite chain is irreducible and aperiodic, so the Markov chain ergodic theorem identifies these probabilities with long-run occupation fractions. The long-run inspected fraction is $1-(1-1/r){\pi }_K=(Kp+q)/(Kp+rq)$. The fraction of all production consisting of undetected defective items is $p(1-1/r){\pi }_K=p(r-1)q/(Kp+rq)$.

An important modeling choice is whether defects during full inspection preserve accumulated good items or reset the count. Resetting produces the distinct consecutive-clearance inspection model and changes the long-run fractions.