Assume $S⊢B_{S}X\to X$ and put $T=S+\{\neg X\}$. Then $T⊢\neg B_{S}X$. The given extension-provability implication, with $Y=\perp$, givesThe provable classical implication $(\neg X\to \perp )\to X$, together with distribution, gives $S⊢B_S(\neg X\to \perp )\to B_{S}X$. HenceIf $T$ were consistent, this would violate its Gödel second incompleteness theorem. Thus $T$ is inconsistent. The deduction theorem for propositional logic and double-negation elimination give $S⊢\neg \neg X$, hence $S⊢X$. Therefore the second incompleteness theorem for $S+\{\neg X\}$ yields Löb's theorem for arbitrary $X$.

Assume $S⊢\neg X$ and $S⊢BX\to X$. Combining these theorems gives $S⊢\neg BX$. Because $X$ is refutable, $\neg BX$ is a legitimate formal consistency statement for $S$: the standard contradiction first-order sentence and $X$ are provably interderivable, and the derivability conditions transfer those implications to their provability statements.

Thus if $S$ were consistent, the Gödel second incompleteness theorem would contradict $S⊢\neg BX$. Consequently the assumed reflection theorem forces $S$ to be inconsistent; explosion then gives $S⊢X$. This establishes the instance of Löb's theoremfor every $X$ refutable in $S$, using only the second incompleteness theorem.

The Gödel second incompleteness theorem says that a consistent, effectively axiomatized first-order theory with enough arithmetic and its standard derivability and diagonalization properties cannot prove its formal consistency statement ${\mathrm{Con}}_S=\neg B\perp$.

If a consistent $S$ proved ${\mathrm{Con}}_S$, it would prove $B\perp \to \perp$ by classical propositional logic. Apply Löb's theorem to $A=\perp$. This would give $S⊢\perp$, contradicting consistency. ThereforeFor an inconsistent first-order theory the conclusion need not hold, since it proves every first-order sentence.

For the surface of revolution,The mixed inner product vanishes, givingTaking $f(u)=\cosh u$ makes both coefficients ${\cosh }^{2}u$, as used in the parent calculation.

Differentiate to getTheir squared lengths are $a^2$ and $a^2{u}^2+1$, and their inner product is zero. Hence the helicoid has first fundamental form

Put $R(z)=\frac{1}{4}({\wp }^{\prime \prime }/{\wp }^{\prime }{)}^2-2\wp (z)$. This is an elliptic function. Its only possible poles modulo $\Lambda$ are $0$ and the three simple zeros $h$ of ${\wp }^{\prime }$. At such a nonzero half-period, periodicity and evenness give $\wp (h+t)=\wp (h-t)$, soIt follows that $R(h+t)=1/(4t^2)+O(1)$ with no simple-pole term. This is exactly the principal part of $\wp (2h+2t)=\wp (2t)$.

At zero the source expansion yieldsConsequently $R(z)=1/(4z^2)+O(z^2)$, again matching $\wp (2z)$. Thus $R(z)-\wp (2z)$ is a holomorphic function on a compact connected Riemann surface, hence constant. Its limit at zero is zero, provingThe equality extends over the apparent singularities as an identity of meromorphic functions.

A pole occurs exactly when $2z\in \Lambda$. Thus the poles on $T$ areEach is a double pole, since the multiplication-by-two map is locally unramified and $\wp$ has a double pole at $[0]$. There are four distinct poles, each of order two, consistent with the degree eight.

Multiplication by two on the complex torus is a holomorphic unramified coveringIts kernel is $(\frac{1}{2}\Lambda )/\Lambda$, of order four; its derivative is the nonzero constant two. Therefore $\mathrm{deg}[2]=4$. Composition with the degree-two Weierstrass elliptic function givesIt is well-defined because changing $z$ by $\omega \in \Lambda$ changes $2z$ by the period $2\omega$.

The Riemann-Hurwitz formula gives total ramification four because $0=2\cdot 2-R$. The double pole at $[0]$ has local degree two and contributes one. Since ${\wp }^{\prime }$ is odd and periodic, it vanishes at each nonzero point satisfying $[h]=[-h]$. These areThey are three distinct finite points, each contributing at least one to ramification. Together with $[0]$ they exhaust the total of four. Each therefore has local degree exactly two, meaning its derivative has a simple zero. There are no other zeros of ${\wp }^{\prime }$, and their translates by $\Lambda$ give the full list in the complex plane.

Let $\Lambda =\mathbb{Z}{\omega }_1+\mathbb{Z}{\omega }_2$ and $T=\mathbb{C}/\Lambda$. Its fundamental parallelogram with opposite sides identified is a torus, so $\chi (T)=0$. The Weierstrass elliptic function is even and doubly periodic. One way to verify the latter from its series is to differentiate: the locally normally convergent derivative series iswhich is absolutely convergent away from the lattice and periodic by reindexing. Thus $\wp (z+{\omega }_j)-\wp (z)$ is constant. Evaluating at $z=-{\omega }_j/2$ and using evenness shows that this constant is zero for each generating period.

Therefore $[z]\mapsto \wp (z)$ descends to a meromorphic function on $T$, with one double pole at $[0]$. Viewed as a map to the Riemann sphere, a pole is a holomorphic point in the target coordinate $1/w$. The map is consequently holomorphic everywhere and

Here the surface is the real projective plane, with Euler characteristic one. The bound gives $V\ge 6$. The following ten triangles on vertices $1,\dots ,6$ attain it:Take these as the maximal simplices of an abstract simplicial complex, identifying edges with the same vertex labels. Every one of the 15 pairs of vertices occurs as an edge in exactly two triangles. The links of the six vertices are single five-cycles; listed cyclically they areThus every point has a disk neighborhood, and the realization is a connected closed surface. Its Euler characteristic is $6-15+10=1$, so the classification of compact surfaces identifies it with the real projective plane. This is an explicit triangulation, proving $\boxed{V_{\min }=6}$.

The simplicial intersection conditions prohibit loops and multiple edges between the same pair of vertices, so the one-skeleton is a simple graph andUsing the preceding identity gives $V^2-7V+6\chi \ge 0$. A closed surface has $\chi \le 2$ and a simplicial triangulation has $V\ge 4$. The smaller root is at most three, so this inequality forces $V$ to lie above the larger root:For completeness, fewer than four vertices could give at most one triangle under the intersection rules, which cannot form a closed surface.

Count incidences of triangular faces with edges. Every triangle contributes three and every edge of a closed surface occurs in two triangles, hence $3F=2E$. Substituting $F=2E/3$ into the Euler characteristic gives

Set $g(w)=a\wp (w)+b{\wp }^{\prime }(w)+c$. The Weierstrass elliptic function is doubly periodic, with a double pole at the origin modulo $\Lambda$ and no other poles. Its derivative has leading term $-2w^{-3}$. Since $b\ne 0$, no lower-order term cancels this, so $g$ has a pole of order exactly three at zero and no others.

The point $\Phi (0+\Lambda )=[0,1,0]$ does not lie on the line, because its line equation evaluates to $b\ne 0$. At every other point of the image, lying on the line is equivalent to $g(w)=0$. The three distinct image points therefore yield three distinct zeros modulo $\Lambda$. By the zero-pole count, these are all the zeros and each is simple. Hence its principal divisor on the torus isApplying the elliptic divisor-sum identity givesThis is the period lattice form of the chord-and-tangent group law on an elliptic cubic.

For a divisor on an algebraic curve $D={\sum }_j{n}_j[p_j+\Lambda ]$, defineChanging the representatives $p_j$ changes the second coordinate only by an element of $\Lambda$. The two preceding contour arguments show that a principal divisor has degree zero and period lattice sum zero. Thus the formula is unchanged by linear equivalence of divisors and is a well-defined group homomorphism on the quotient.

It is nontrivial, since $[0]$ maps to $(1,0+\Lambda )$. In fact it is surjective: for any $d\in \mathbb{Z}$ and $z+\Lambda \in \mathbb{C}/\Lambda$, the divisor on an algebraic curve $(d-1)[0]+[z+\Lambda ]$ maps to $(d,z+\Lambda )$. Only well-definedness and a nontrivial homomorphism are required here; identifying its entire kernel would be an additional theorem.

Order the period lattice basis so the indicated contour is positively oriented; reversing its orientation only changes the signs of the resulting integers. SetPairing the translated sides while retaining the factor $w$ givesFor example, the right side is the translate by ${\lambda }_1$ of the left side with reversed orientation; their combined contribution is ${\lambda }_{1}B$. The bottom and top similarly give $-{\lambda }_{2}A$.

Along either path there is a continuously continued logarithm of the nonzero finite function $f$. Since its endpoint values agree by periodicity, each logarithmic increment is an integer multiple of $2\pi i$. Therefore $A=2\pi ir$, $B=2\pi is$ for integers $r,s$. The residue theorem applied to $wh$ now yieldsThis is the elliptic divisor-sum identity.

First record the unlettered definitions. A divisor on an algebraic curve on a compact Riemann surface is a finite formal sum $D={\sum }_p{n}_p[p]$ with $n_p\in \mathbb{Z}$. Its degree is $\mathrm{deg}D={\sum }_p{n}_p$. A principal divisor is $(f)={\sum }_p{\mathrm{ord}}_p(f)[p]$ for a nonzero meromorphic function, with positive order at a zero and negative order at a pole. Two divisors are linearly equivalent if their difference is principal. Principal divisors form a subgroup because $(fg)=(f)+(g)$ and $(1/f)=-(f)$.

For the doubly periodic function, put $h=f^{\prime }/f$. It is periodic with the same period lattice, and its residue at $p_i$ equals $n_i$. Opposite sides of the fundamental parallelogram are translates by a period with opposite directions, so their integrals of $h$ cancel. The residue theorem givesThusIn particular an elliptic function has equally many zeros and poles counted with multiplicity in a period cell.

Let the six distinct conic points be $p_1,\dots ,p_6$, and write ${\ell }_{ij}$ for their joining line. Form two reducible cubicsNo three distinct points of an irreducible conic are collinear, by the Bézout theorem applied to a line and the conic. Thus all six side lines are distinct, the cubics have no common component, and six of their intersections are precisely the vertices. The remaining pairings giveThese three points are off the conic and distinct. For example, equality of $q_1$ and $q_2$ would put that point on both ${\ell }_{34}$ and ${\ell }_{45}$, making it $p_4$ and then placing $p_4$ on ${\ell }_{12}$, impossible. The other cases are identical. Hence the two cubics meet in nine distinct points.

Apply residual intersection through a curve pencil with $n=3$, $m=2$ and $E$ the conic: six intersections lie on $E$, so the other three lie on a curve of degree at most one. Therefore $q_1,q_2,q_3$ are collinear, proving Pascal's theorem.

The curves have no common component, since their intersection is finite. Let $R=0$ define the irreducible degree-$m$ curve $E$. Choose $p\in E$ outside $C\cap D$, and select $(\lambda ,\mu )\ne (0,0)$ with $\lambda P(p)+\mu Q(p)=0$. The polynomial $H=\lambda P+\mu Q$ is not identically zero, since $P,Q$ are not proportional.

The degree-$n$ curve $H=0$ meets $E$ at the prescribed $nm$ base points and also at $p$. If $R$ did not divide $H$, irreducibility of $R$ and the Bézout theorem would bound the number of distinct intersections by $nm$, contradicting these $nm+1$ points. ThusAt every remaining common zero of $P,Q$ lying off $E$, $R$ is nonzero and hence $S=0$. ThereforeThis is residual intersection through a curve pencil. The existence of a point $p$ outside the finite base locus is automatic on the projective curve $E$.

Take the union of $n$ distinct projective lines in general position, with no three concurrent. Its defining polynomial is the product of their linear equations, so its degree is $n$. Exactly two lines meet at each of the $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ pairwise intersections, producing an ordinary double point; elsewhere just one line passes through and the curve is smooth. HenceFor $n=1$, the line itself gives zero singular points, as required.