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Let $v_1\cdots v_r$ be a longest path in the finite simple graph. Every neighbour of either endpoint lies on it. Among indices $1\le i<r$, consider those with edges $v_1{v}_{i+1}$ and those with edges $v_i{v}_r$. Their cardinalities are the endpoint degrees, whose sum is at least $n\ge r$, exceeding the $r-1$ available indices. They intersect, producing the cycleEvery connected component has at least $\delta (G)+1>n/2$ vertices, so the graph is connected. If $r<n$, an edge joins a vertex outside this cycle to it, and breaking the cycle there gives a longer path, a contradiction. Hence $r=n$ and this is a Hamilton cycle, proving Dirac theorem.

The complete graph $K_4$ is planar, has chromatic number four and a Hamilton cycle. Adding one leaf attached to one of its vertices preserves planarity and chromatic number four but prevents a Hamilton cycle, since a leaf has degree one.

For the last claim, the given plane embedding has all vertices on the outer cycle. Add noncrossing diagonals until its polygonal interior is triangulated. A triangulated polygon has a vertex incident with only its two boundary neighbours: for example the dual graph of its triangles is a tree and a leaf triangle supplies an ear. Remove this vertex. Inductively three-colour the smaller triangulated polygon, starting with a triangle. The removed vertex's neighbours are adjacent and have different colours, so it can receive the third. Restriction to the original graph givesThis uses the specified embedding, not the false assertion that every planar Hamiltonian graph is three-colourable; $K_4$ is already a counterexample to that broader assertion.

The completeness theorem for propositional logic, together with soundness, states $S⊢\phi$ if and only if $S\models \phi$. In particular, an inconsistent set of formulae has a finite formal proof of contradiction. Such a proof uses only finitely many assumptions, yielding the propositional compactness theorem: if every finite subset of $S$ is satisfiable, then $S$ is satisfiable.

The decidability theorem for propositional logic concerns theoremhood of a formula, or entailment from a finite list of premises. Formulae involve finitely many variables, so their valuations can be tested by a finite truth table; completeness identifies semantic validity with provability. Equivalently, dovetail effective proof enumeration with the finite search for a countervaluation: completeness ensures that one succeeds. This is not a decidability claim for consequences of an arbitrary noncomputable infinite theory.

An infinite collection of distinct tautologies is a finitary set of formulae, since its deductive closure equals that of the empty set. By contrast, the set of independent variables $\{p_i:i\ge 1\}$ is not finitary. If a finite set $T$ had the same closure, it would mention only finitely many variables. It is satisfiable because it is a consequence of the all-true model of the $p_i$; changing an unmentioned $p_j$ to false preserves $T$ and contradicts $T\models p_j$.

Under the final hypothesis, $A\cup B$ has no model. The propositional compactness theorem gives finite $A^{\prime }\subset A$, $B^{\prime }\subset B$ with $A^{\prime }\cup B^{\prime }$ inconsistent. Any valuation satisfying $A^{\prime }$ cannot satisfy $B$, because it would then satisfy $B^{\prime }$; by the exactly-one hypothesis it must satisfy $A$. The reverse implication holds by inclusion. Thus $A$ and $A^{\prime }$ have identical models, and similarly $B$ and $B^{\prime }$. Completeness gives identical deductive closures, proving

Differentiate the Friedmann equation and use energy conservation to obtain $\dot{H}=-4\pi G(\rho +P/c^2)$ on the expanding branch, where $H\ne 0$. Since $\ddot{a}/a=\dot{H}+H^2$,For pressureless matter, conservation gives ${\rho }_M={\rho }_{M0}{a}^{-3}$. In terms of the specified cosmological density parameters, flatness implies ${\Omega }_M+{\Omega }_{\Lambda }=1$, andSet $b=a^{3/2}$ and take ${\Omega }_M>0$. Then $\dot{b}=\frac{3}{2}{H}_0\sqrt{{\Omega }_M+{\Omega }_{\Lambda }{b}^2}$. Integrating with the Big Bang at $t=0$ givesFor another time origin replace $t$ by $t-t_B$. The cosmological constant has effective density ${\rho }_{\Lambda }=3H_0^2{\Omega }_{\Lambda }/(8\pi G)$. Equality with the matter density means $a^3={\Omega }_M/{\Omega }_{\Lambda }$, soIf ${\Omega }_M=0$, the expanding solution is pure exponential de Sitter scale factor in flat slicing and there is no finite matter-density equality time or dust Big Bang; the stated hyperbolic-sine formula presumes a nonzero dust component.

For $\mathrm{Re}s>1$, expand $(e^t-1{)}^{-1}={\sum }_{n\ge 1}{e}^{-nt}$. Absolute integrability, bounded near zero by a constant times $t^{\mathrm{Re}s-2}$, permits termwise integration. The Gamma function integral then givesSpecify the Hankel contour to run from the negative real axis below the cut towards zero, circle zero counterclockwise with fixed radius $0<r<2\pi$, and return above the cut. Take $-\pi <\arg t<\pi$. The circle excludes every nonzero pole $2\pi i\mathbb{Z}$ of the denominator. For $\mathrm{Re}s>1$ its radius can shrink to zero, and the lower and upper rays give respectively $-e^{-i\pi s}\Gamma (s){\zeta }_R(s)$ and $e^{i\pi s}\Gamma (s){\zeta }_R(s)$. By the Gamma reflection formula,away from integer $s$ in that half-plane, with limiting equality at the removable points.

With the circle radius kept fixed, the contour integral is an entire function of $s$: the rays decay exponentially and converge locally uniformly, and the circular part stays away from zero. Multiplication by $\Gamma (1-s)$ is analytic for $\mathrm{Re}s<1$ and meromorphic globally. The apparent singularities at positive integers $s\ge 2$ are removable because the integral vanishes there and it already agrees with the original analytic zeta function nearby. Only $s=1$ remains a simple pole. Thus this constructs the meromorphic continuation to the whole plane and an analytic continuation on $\mathbb{C}\setminus \{1\}$. Keeping the circular segment is essential when the integral is not locally integrable at zero.

At $s=-1$ the two rays cancel, since $t^{-2}$ is single-valued. The Laurent expansionshows that the residue of $t^{-2}/(e^{-t}-1)$ is $-1/12$. Since $\Gamma (2)=1$, the residue theorem gives

Under the binomial distribution model's assumption of independent games with a common win probability $p$, the probability of winning a match isThis is also the cumulative distribution function $F_{6,6}(p)$ of a beta distribution with parameters $(6,6)$: differentiation of the binomial tail gives $h^{\prime }(p)=11!p^5(1-p{)}^5/(5!{)}^2$, and $h(0)=0$. In particular $h$ is strictly increasing on $(0,1)$ and maps $[0,1]$ bijectively onto itself.

The new match probability is ${\tilde{P}}_{ab}=h(P_{ab})$, while the desired linear predictor remains the game-level log odds. Therefore the link function isIts endpoint values are $-\infty$ and $+\infty$. This is not the ordinary match-level log odds link: it first recovers the underlying game probability. Symmetry of the binomial tail gives $h(1-p)=1-h(p)$, hence $g(1-u)=-g(u)$, as required by swapping the players. Dependence or varying game probabilities within a match would invalidate the stated binomial-tail conversion.

Let $S_0=0$ and let each increment of the simple symmetric random walk be uniformly chosen from $\{\pm e_1,\pm e_2,\pm e_3\}$. A return requires an even number of steps. In $2n$ steps returning to zero, let $a,b,c$ be the respective numbers of positive steps in the three directions; there must also be $a,b,c$ negative steps, so $a+b+c=n$. Counting step sequences givesThe terms $p_{abc}=n!/(3^{n}a!b!c!)$ form a multinomial distribution, so $\sum p_{abc}^2\le (\max p_{abc})\sum p_{abc}=\max p_{abc}$. The permitted combinatorial inequalities give $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/4^n\le C_1{n}^{-1/2}$ and ${\max }_{a+b+c=n}{p}_{abc}\le C_2/n$ for $n\ge 1$. The second bound also follows by noting that the largest multinomial coefficient occurs when $a,b,c$ differ by at most one and then applying Stirling formula. ThusTo prove the origin is a transient state, let $q$ be the probability of at least one return after time zero. The Strong Markov property at successive returns gives $\mathbb{P}(N\ge k)=q^k$ for the number $N$ of later returns. Therefore $\mathbb{E}(N+1)={\sum }_j{u}_j$ would be infinite if $q=1$. The bound forces $q<1$, proving that the origin is a transient state. Translation invariance gives the same conclusion at every lattice point. The coordinate walks need not be treated as independent; the counting via multinomial coefficients already incorporates their dependence.

The successive differences of the constant entries give a simple LDL decomposition:For example, entry $(i,j)$ of $LD{L}^T$ is the sum of the first $\min (i,j)$ entries of $D$, directly verifying every entry of $A$. Because $L$ is an invertible matrix, $x^{T}Ax=(L^{T}x{)}^{T}D(L^{T}x)$ is strictly positive for every nonzero vector exactly when every diagonal entry of $D$ is positive. ConsequentlyAt $\lambda =14$ it is a positive semidefinite matrix and singular; for $\lambda <14$ its quadratic form takes both positive and negative values.

When $\lambda =30$, taking positive square roots of $D$ yields the Cholesky decomposition

Choose an origin near the bounded loop and write ${\mathbf{r}}^{\prime }$ for the source position. The magnetic vector potential and its far-field multipole expansion arewith the loop held fixed as $r=|\mathbf{r}|\to \infty$. The first term integrates to zero since $C$ is closed. Define its oriented vector area byThe second equality is Stokes theorem, applied componentwise to any oriented spanning surface. In particular the vector area is independent of the choice of spanning surface. From ${\oint }_{C}d(r_i^{\prime }{r}_j^{\prime })=0$, the matrix ${\oint }_C{r}_i^{\prime }d{r}_j^{\prime }$ is antisymmetric, which gives ${\oint }_C(\mathbf{r}\cdot {\mathbf{r}}^{\prime })d{\mathbf{r}}^{\prime }=\mathbf{S}\times \mathbf{r}$. HenceThe orientation follows the current by the right-hand rule. For $r\ne 0$, taking the curl of this leading magnetic vector potential gives the magnetic dipole fieldFor a planar loop, $\mathbf{S}$ is the signed area times the unit normal vector. For a nonplanar loop it is the oriented vector area, rather than the scalar area of an arbitrarily chosen surface.

Choose a star centre $a$ of the open star-shaped set $D$ and define the candidate antiderivative by a straight-segment contour integral:For each fixed $z\in D$ and all sufficiently small $h$, the filled triangle with vertices $a,z,z+h$ lies in $D$. Indeed, the compact segment $[a,z]$ has a neighbourhood of some positive radius contained in the open set $D$, and every point of this triangle lies within $|h|$ of that segment. The zero triangle integral consequently givesBy continuity, the last integral tends to $f(z)$ as $h\to 0$. HenceIn particular, this proves the existence of the antiderivative without incorrectly assuming the star-shaped set is convex.

On a general domain, the conclusion can fail. Take $D=\mathbb{C}\setminus \{0\}$ and $f(z)=1/z$. Every filled triangle contained in $D$ has zero boundary integral by Cauchy integral theorem, but the unit circle has integral $2\pi i$. An antiderivative would make every closed contour integral zero. This is the period obstruction to a holomorphic antiderivative: the local conclusion of Morera's theorem does not remove the global obstruction.