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Löb's theorem states that, for a first-order theory with the stated provability and diagonalization properties,where $BA=P_S(┌A┐)$. It does not require an extra consistency assumption.

By the diagonal lemma, choose $D$ with $S⊢D\leftrightarrow (BD\to A)$. necessitation rule and distribution applied to $D\to (BD\to A)$ giveUsing positive introspection $BD\to B(BD)$ gives the unconditional theorem $S⊢BD\to BA$. If $S⊢BA\to A$, it follows that $S⊢BD\to A$, and the reverse fixed-point implication gives $S⊢D$. Necessitating gives $S⊢BD$, hence $S⊢BA$ and finally $S⊢A$.

The same derivation also gives the Internal form of Löb's theorem, needed for modal soundness. Put $H=(BA\to A)$. From the unconditional $BD\to BA$, propositional reasoning and the fixed point give $S⊢H\to D$. necessitation rule and distribution imply $S⊢BH\to BD$, and henceThis internal statement was derived without necessitating a temporary assumption.

Let $T=S+\{{\mathrm{Con}}_S\}$ and assume $T$ is consistent. The allowed arithmetizations of the previous equivalences giveSuppose $S⊢R\to G$. Its contrapositive is $\neg G\to \neg R$, and the Hilbert-Bernays-Löb derivability conditions then giveCombining with the two internal equivalences yields $S⊢{\mathrm{Con}}_S\to \mathrm{Con}(T)$. Since $T$ contains ${\mathrm{Con}}_S$, it would prove $\mathrm{Con}(T)$, contrary to the given Gödel second incompleteness theorem for this consistent extension. HenceTogether with the preceding implication, this is the Strict Gödel-to-Rosser implication under the stated consistency hypothesis.

The preceding Gödel sentence argument gives $S⊢G\to {\mathrm{Con}}_S$. The permitted arithmetized positive half of the Rosser sentence argument isPure first-order logic gives $\neg BR\to R$: if there is no proof of $R$, each antecedent in its defining universal implication is false. ThereforeThis uses the supplied internal Rosser implication, rather than replacing an external nonprovability fact by an unproved internal first-order sentence.

If $S$ is inconsistent, it proves $\neg R$ by explosion. Conversely, suppose $S$ is consistent and, towards a contradiction, has a proof of $\neg R$ with standard code $n$. Consistency excludes any proof of $R$. In particular, for each $k\le n$, the closed bounded statement $\neg \mathrm{Prov}(┌R┐,\overline{k})$ is true and is provable in $S$ by Sigma-1 completeness. The true bounded statement $\mathrm{Prov}(┌\neg R┐,\overline{n})$ is also provable.

Using the given finite-numeral exhaustion below $\overline{n}$, these finitely many nonproof facts giveFor arbitrary $y$, the given comparison splits into $y\le \overline{n}$ or $\overline{n}\le y$. In the first case the antecedent $\mathrm{Prov}(┌R┐,y)$ is false. In the second case the known negative proof at code $\overline{n}$ supplies the required witness $z\le y$. Hence $S$ provesThe Rosser sentence equivalence now gives $S⊢R$, contradicting consistency. ThusOnly a fixed finite list of closed bounded facts was used; no assumption that the first-order theory is omega-consistent is needed.

Write $BA=\exists y\mathrm{Prov}(┌A┐,y)$ and fix a refutable first-order sentence $\perp$, so $S⊢\neg \perp$ and ${\mathrm{Con}}_S=\neg B\perp$. First prove the internal equivalenceBecause $\perp \to G$ is provable, the derivability conditions give $B\perp \to BG$. The fixed-point equivalence $G\leftrightarrow \neg BG$ therefore yields $G\to \neg B\perp$.

For the reverse implication, necessitating $G\to \neg BG$ and applying distribution gives $BG\to B(\neg BG)$. positive introspection gives $BG\to B(BG)$. Finally the provable implication $BG\to (\neg BG\to \perp )$, necessitated and distributed, givesConsequently $S⊢BG\to B\perp$, so $\neg B\perp \to \neg BG\to G$. The same argument works with any chosen refutable first-order sentence in place of $\perp$; their formal consistency statements are provably equivalent.

By the classical deduction theorem for propositional logic applied to adjoining a first-order sentence, $S+\{G\}$ is consistent exactly when $S⊬\neg G$. The displayed equivalence makes $S+\{G\}$ and $S+\{{\mathrm{Con}}_S\}$ deductively equivalent. ThereforeNo soundness assumption on that extension is being substituted for consistency.

Write $BA=P(┌A┐)$, with the corner notation understood as the numeral for the Gödel number. An arithmetical provability predicate numerates the theorems of $S$ in the standard natural numbers and satisfies the Hilbert-Bernays-Löb derivability conditions:The last two conditions are internal theorems, not rules authorizing reflection from $BA$ to $A$.

Here $P(x)=\exists y\mathrm{Prov}(x,y)$ numerates the theorem set by the given assumption. If $S⊢A$, its proof supplies a genuine natural-number witness, making $BA$ a true Sigma-1 formula. Sigma-1 completeness gives $S⊢BA$, proving the first condition. The second condition is supplied explicitly. Since $BA$ is itself a Sigma-1 formula, the given formalized Sigma-1 completeness yields $S⊢BA\to B(BA)$, the third condition. Thus the specified existential proof predicate is an arithmetical provability predicate.

Given $f$, choose the hyperplane divisors as in (a). The ratios $f_Q$ form a linear subspace of $L(D)$ by (d), and its dimension equals $md+1-g=\ell (D)$ by (b). A linear subspace of a finite-dimensional vector space with the same dimension is the whole space. Hence $f=f_Q$ for some homogeneous polynomial $Q$ of degree $m$, givingThe zero function corresponds to $Q=0$. Thus every allowed meromorphic function is a ratio of homogeneous polynomials of equal degree, with a denominator that can be chosen as a product of linear forms.

Write $L_i=a_{i}x+b_{i}y+c_{i}z$ and $S={\prod }_i{L}_i$. The numerator and denominator have the same homogeneous degree, so their ratio is well-defined on projective points away from its poles and extends as a meromorphic function on $C$. Locally the pole order cannot exceed the zero order of $S{|}_C$, whose divisor is $D$. Thus $f_Q\in L(D)$.

Equality of two ratios is equivalent to $Q-Q^{\prime }$ vanishing on $C$. A nonsingular projective plane curve is irreducible: distinct positive-degree components would intersect by Bézout's theorem, producing a singular point. Its homogeneous vanishing ideal is therefore generated by its irreducible defining polynomial $P$. One can also see this using Bézout's theorem: a homogeneous polynomial not divisible by $P$ has only finitely many intersections with $C$, so cannot vanish on the whole curve. ConsequentlyMultiplication by the nonzero polynomial $P$ is injective. The linear map $V_m\to L(D)$ therefore has kernel $P{V}_{m-d}$ and image dimensionThe assumption $m\ge d$ makes both polynomial-space dimensions applicable directly.

The monomials $x^i{y}^j{z}^k$ with nonnegative integers $i+j+k=m$ form a basis for the vector space of homogeneous polynomials of degree $m$. For each $i=0,\dots ,m$, there are $m-i+1$ choices of $(j,k)$. Hence

The divisor $D={\sum }_i{H}_i$ has degree $md$. The hypothesis gives $\mathrm{deg}(K-D)=2g-2-md<0$. A divisor of negative degree has no nonzero section: if $h\in L(K-D)\setminus \{0\}$, then $(h)+K-D$ would be effective, but its degree would be negative because the divisor of a meromorphic function has degree zero. Thus $\ell (K-D)=0$. The Riemann-Roch theorem now gives

A meromorphic function on a compact Riemann surface has finitely many poles, say $p_j$ of orders $r_j$. For each $p_j$, choose a projective line through it that does not contain $C$. Its hyperplane divisor contains $p_j$ with multiplicity at least one. Include $r_j$ copies of this divisor. The sum then dominates the pole divisor of $f$.

Add further arbitrary hyperplane divisors until their number $m$ satisfies $m\ge d$ and $md>2g-2$. Adding effective divisors cannot destroy the pole bound, so for $D=H_1+\cdots +H_m$ we haveFor a constant function, including zero, no initial poles need to be covered; simply add sufficiently many lines.

Take $\lambda =e^{\pi i/3}$ and the chosen $s=e^{\pi i/6}$. Then $\lambda s=i$, so $i/(\lambda s)=1$, and $-1/\lambda ={\lambda }^2$. In affine coordinates the chosen lift becomesSince ${\lambda }^2$ is a primitive cube root of unity, $1+{\lambda }^2+{\lambda }^4=0$. Three iterations fix $x$ and $y$, whereas one does not. ThusThe other square-root choice replaces this map by $(x,y)\mapsto (1+{\lambda }^{2}x,-y)$. Its cube is $(x,y)\mapsto (x,-y)$, so that lift has order six. This sign dependence is why a lift was specified explicitly.

Both fix the point at infinity. In the complex torus model their linear multipliers have order three or six, so the lattice is stable under a primitive cube root $\zeta =e^{2\pi i/3}$ and hence under $e^{\pi i/3}=1+\zeta$. Choose a shortest nonzero period $\omega$. The triangular lattice $\mathbb{Z}+\zeta \mathbb{Z}$ has covering radius $1/\sqrt{3}<1$: its equilateral triangles of side one each have circumradius $1/\sqrt{3}$. Approximating any period divided by $\omega$ by this lattice leaves a strictly shorter period, which must be zero. ThereforeThus the cubic has the hexagonal lattice shape, independently of which of the two lifts is chosen.

At $\lambda =-1$ and $s=i$, the chosen map isThe square is not the identity on the cubic, since a generic point has $y\ne 0$. Thus $\boxed{\mathrm{ord}(\sigma )=4}$.

Choose the cubic's point $[0:1:0]$ as the origin in its complex torus description. Since $\sigma$ fixes it, its torus form is $z+\Lambda \mapsto az+\Lambda$, with $a\Lambda =\Lambda$. Exact order four implies $a=\pm i$; $a^2=-1$. Therefore $\Lambda$ is stable under multiplication by $i$.

To identify its shape without an unproved uniqueness assertion, choose a shortest nonzero period $\omega$. For any ${\lambda }_0\in \Lambda$, approximate ${\lambda }_0/\omega$ by a Gaussian integer $m+ni$ within distance $1/\sqrt{2}<1$. The difference ${\lambda }_0-(m+ni)\omega$ belongs to $\Lambda$ and is shorter than $\omega$, so it must vanish. HenceThe lattice is determined up to a nonzero complex scaling: it is square.

On the parameter, the two transformations are $S(\lambda )=1/\lambda$ and $T(\lambda )=1-\lambda$. Their compositions yieldEvery step lifts to an invertible projective transformation of the corresponding cubic curves. Therefore all six parameter values give isomorphic curves. Some values coincide for special $\lambda$; this does not affect the conclusion.

Compose the preceding transformations, applying ${\sigma }_{\lambda }$ first and then ${\tau }_{1/\lambda }$. With the square-root convention already specified,This invertible projective transformation fixes $[0:1:0]$ and carries $C_{\lambda }$ to $C_{1-1/\lambda }$. It is important to retain the sign of the chosen lift on the $y$ coordinate when computing its order later.

An appropriate projective transformation isIt fixes $[0:1:0]$ and has an invertible representing matrix. For $X=z-x$, $Y=iy$, $Z=z$,Hence its restriction is an isomorphism $C_{\lambda }\to C_{1-\lambda }$.

Choose $s\in \mathbb{C}$ with $s^2=\lambda$. Since $\lambda \ne 0$, the invertible diagonal projective transformationfixes $[0:1:0]$. Writing its image as $[X:Y:Z]$, direct substitution givesThus it maps $C_{\lambda }$ onto $C_{1/\lambda }$. Either choice of square root works. We choose $s=i$ at $\lambda =-1$ and $s=e^{\pi i/6}$ at $\lambda =e^{\pi i/3}$ for the later order calculations.

Injectivity makes the images of any three distinct points distinct. By the ordered-frame result, there is a projective transformation $L$ agreeing with the given map on three chosen points. Composing by $L^{-1}$ reduces to an injective map fixing those points and preserving every cross-ratio. Change coordinates so that the three points are $\infty ,0,1$.

For any other point with affine coordinate $z$, the convention in the source givesPreservation therefore forces the image of $z$ to have the same coordinate $z$. The normalized map fixes every point and is the identity. Undoing the normalization proves the original map is exactly $L$, hence projective. No prior continuity or surjectivity assumption is needed.

Represent each point by a nonzero column $v_i$. The source's cross-ratio isRescaling representatives cancels between numerator and denominator. If the projective transformation is represented by $A\in {\mathrm{GL}}_2(\mathbb{C})$, then $\det (A{v}_i,A{v}_j)=\det (A)\det (v_i,v_j)$. Two factors of $\det (A)$ occur in both numerator and denominator and cancel. Hence the cross-ratio is invariant under every projective transformation.

Choose representatives $v_1,\dots ,v_{n+1}$ of the first $n+1$ points. The general-position condition says that these vectors form a basis of ${\mathbb{C}}^{n+1}$. Write a representative of the last point as $v_{n+2}={\sum }_i{c}_i{v}_i$. Every $c_i$ is nonzero; otherwise the last point and the other $n$ basis points would lie in a hyperplane. Rescaling $v_i$ by $c_i$ gives representatives with $v_{n+2}={\sum }_i{v}_i$. Normalize representatives $w_i$ of the second frame in the same way.

There is a unique invertible linear map $L$ with $L(v_i)=w_i$ for $i\le n+1$. It also maps $v_{n+2}$ to $w_{n+2}$, so its projectivization has the required effect. For uniqueness, the quotient of two candidate projective transformations fixes the normalized frame. A representing matrix must be diagonal in the $v_i$ basis, with entries $a_i$. Fixing $[{\sum }_i{v}_i]$ forces all $a_i$ to be equal. Such a scalar represents the identity projective transformation. Thus the ordered projective frame determines the transformation uniquely.