Consider extending $S+{\mathrm{Con}}_S$ by a rule sending every ${\mathrm{Con}}_A$ to ${\mathrm{Con}}_{A+{\mathrm{Con}}_A}$. Put $A=S+\neg {\mathrm{Con}}_S$. The initial consistency axiom proves ${\mathrm{Con}}_A$ by the negated-consistency extension theorem, so the rule proves ${\mathrm{Con}}_{A+{\mathrm{Con}}_A}$. But ${\mathrm{Con}}_A\to {\mathrm{Con}}_S$ is provable by inclusion of proofs, making $A+{\mathrm{Con}}_A$ inconsistent. Arithmetic verifies its finite contradiction proof. Thus the unrestricted rule proves both that theory's consistency and inconsistency. This differs from an externally justified progression along specified sound theories.

For a formal group law over any commutative ring $R$, the invariant differential of a formal group law satisfies $[p{]}^{\ast }\omega =p\omega$. Its coefficient is a unit in a ring, so $[p{]}^{\prime }(T)\in pR[[T]]$. If $[p](T)=\sum a_m{T}^m$, this means $m{a}_m\in pR$. When $p\nmid m$, the integer Bézout's identity implies $a_m\in pR$, even if $p$ is a zero divisor. Put those coefficients into $pf(T)$ and the exponents divisible by $p$ into $g(T^p)$.

A regular function $f$ on an entire affine variety is a polynomial function, even if the variety is reducible. Put $J=\{g\in A(X):gf\in A(X)\}$, an ideal of its coordinate ring. Near a point $p$, write $f=a/b$ with $b$ nonzero, and choose a principal open subset $D(h)$ containing $p$ inside that neighbourhood. Then $(hb)f=ha$ on all of $X$: on $D(h)$ this is the fraction identity, and off $D(h)$ both sides vanish. Hence $hb\in J$ and $(hb)(p)\ne 0$. No point is a common zero of $J$, so the weak Hilbert Nullstellensatz implies $J=A(X)$. Thus $1\in J$ and $f\in A(X)$. This proof does not require a fraction field for a reducible coordinate ring.

A propositional formula false on the all-true and all-false Boolean valuations has a disjunctive normal form in which every minterm contains a positive literal and a negative literal. Grouping the positive propositional variables into $C$ and the negated propositional variables into the negation of a logical disjunction $D$ writes each minterm as $\neg (C\to D)$. De Morgan's laws therefore give the negation of a logical conjunction of such logical implications. For a contradiction use $\neg (p\to p)$.

A Boolean formula in conjunctive normal form with at most two Boolean literals per Boolean clause. Shorter Boolean clauses are allowed. The Boolean satisfiability problem restricted to these Boolean formulas is 2-SAT.

A Boolean formula in conjunctive normal form with at most three Boolean literals per Boolean clause. A shorter nonempty Boolean clause can be padded by repeating a Boolean literal. The corresponding Boolean satisfiability problem is 3-SAT.

Let $A_t=\exp (t^{-1}{\int }_0^t\log S_{u}du)$ for a positive geometric Brownian motion in the non-dividend-paying Black-Scholes model, and put $\tau =T-t$. Under the risk-neutral measure, the future log-average isThe final integral is a centred normal variable independent of the past, with variance ${\tau }^3/3$. Applying the moment-generating function of a normal distribution proves the displayed identity for every real $p$. Multiplication by $e^{-r\tau }$ prices the payoff $A_T^p$. The historical average must remain a state variable; terminal stock price alone does not determine this payoff.

Let $S_0=0$ and let the independent and identically distributed random variables $S_n-S_{n-1}$ be $+1$ with probability $p>1/2$ and $-1$ with probability $q=1-p>0$. The strong law of large numbers gives $S_n\to +\infty$ almost surely, so the global minimum $W$ is finite and attained. For each nonnegative integer $m$, the unit-step property and translation of the starting point giveby the biased random walk hitting probability. Subtracting consecutive tails gives the displayed zero-based geometric distribution, with success parameter $1-q/p$. Inclusion of time zero is important: it ensures $-W\ge 0$. Reflecting the sign of the path yields the corresponding maximum of a downward-biased random walk.

For $n\ge 3$, let $S_n$ permute the coordinates of $M=F^n$. Write $T$ for the constant vectors and $U=S^{(n-1,1)}$ for the augmentation submodule. These and zero and $M$ are all the submodules. Indeed a vector with two unequal coordinates yields a nonzero multiple of $e_i-e_j$ by subtracting its image under $(i,j)$; the orbit of such a difference spans $U$. A submodule without such a vector lies in $T$. If the characteristic does not divide $n$, then $M=T\oplus U$ and $U$ is simple. If it divides $n$, the lattice is the chain $0<T<U<M$, with $U/T$ simple of dimension $n-2$ and trivial end factors. The direct-sum projection in the first case is $x\mapsto x-({\sum }_i{x}_i)1/n$.

For a nonvanishing kernel tending to the same nonzero value at both ends, the index counts its winding as the contour is traversed from left to right. Index zero permits a continuous logarithm tending to the same normalization at both ends. Canonical bounded Wiener-Hopf factorization and a constant common entire remainder require this qualification. For example $F(z)=(z+i)/(z-i)$ has index $-1$ on the real line; the bounded half-plane functions $w_{+}=(z-i)/(z+i)$ and $w_{-}=1$ obey $F{w}_{+}=w_{-}$. The analytic factors $F_{+}=z+i$, $F_{-}=z-i$ make their common product $z-i$, demonstrating why bounded input functions alone do not guarantee a constant factored remainder.