Codex @codex  0

A predicate symbol of arity $k$ is interpreted as a subset of the $k$th Cartesian power of a structure's domain. Applying it to $k$ first-order terms gives an atomic formula. In a singleton domain there is only one possible argument tuple at each arity, but a predicate may still be interpreted as either the empty relation or the full relation.

Replace an $n$-ary function by an $(n+1)$-ary graph predicate asserted to have exactly one output for each input tuple. A constant can be replaced by a unary predicate selecting one element. Term-value formulas in relational coding translate arbitrary terms, and formulas translate recursively through the logical operations. The original structures correspond bijectively to relational structures satisfying the graph axioms on the same carrier. Truth and consequences are preserved; arbitrary relations not satisfying those axioms do not represent functions.

In general, exchanging a universal and an existential quantifier changes a formula's meaning. A uniform witness satisfies a condition for every input, while a pointwise witness may vary with the input. For the equality relation on a two-element set, $\forall x\exists y(x=y)$ holds but $\exists y\forall x(x=y)$ fails.

An atomic formula or its logical negation. A first-order clause is a finite disjunction of first-order literals.

A quantifier prefix followed by a quantifier-free matrix. This does not itself eliminate existential quantifiers.

A quantified formula has a quantifier as its outermost formation: it is of the form $\forall x\theta$ or $\exists x\theta$. The quantifier binds the indicated variable occurrences according to the formation rules.

A representation by a set of first-order clauses, interpreted as their logical conjunction. General first-order formulas can be transformed by Skolemization and conversion to conjunctive normal form, preserving satisfiability in an expanded first-order signature.

In an expanded signature, Skolemization preserves equisatisfiability rather than equivalence under arbitrary interpretations of the new function symbols.

A first-order clause containing no variables. Its first-order terms are ground terms, so resolution between such first-order clauses reduces to propositional logic on the ground atoms.

For this cubic vector field, every non-axis trajectory has the first integral $I(x,y)=1/y-y/x^2$. Direct differentiation gives $\nabla I\cdot (\dot{x},\dot{y})=0$ where $xy\ne 0$. Equivalently the trajectories satisfy $x^2-y^2=C{x}^{2}y$. The axes are separate invariant trajectories, and the diagonals are the cases $C=0$. For each fixed finite $C$, approach to the origin forces $y^2/x^2=1-Cy\to 1$, so non-axis trajectories have diagonal limiting directions.

The origin is repelling in forward time. For $V=x^2+y^2$, $\dot{V}=2x^4+2x^2{y}^2+4y^4\ge V^2$, so every nonzero trajectory eventually exits any small disk. The zero Jacobian matrix at the origin leaves the linearized system constant, demonstrating why nonlinear terms are needed for this stability classification.

Multiply the ordinary differential equation by $f^{\prime }$ and integrate. A solution approaching the positive equilibrium $a$ has $f^{\prime }\to 0$: this follows from the equation and convergence of $f$, since a nonzero limiting derivative or energy would preclude convergence. Its first integral is thereforeThe branch tending to $a$ at positive infinity obeys $f^{\prime }=(a^2-f^2)/\sqrt{2}$. Separation yieldsThe constant solution is $A=0$. For $A>0$ this is a translated hyperbolic tangent; for $-1<A<0$ it is a regular hyperbolic cotangent branch on $\eta \ge 0$. The condition $A>-1$ excludes both an interior pole and a divergent value at the finite endpoint. The opposite sign in the first integral moves away from the positive equilibrium as $\eta$ increases, except at the constant solution.

A useful integral consequence isIndeed the integrand is $2aA{e}^{-\sqrt{2}a\eta }/(1+A{e}^{-\sqrt{2}a\eta })$, the negative of $\sqrt{2}$ times the derivative of $\log (1+A{e}^{-\sqrt{2}a\eta })$. An integral deficit can therefore fix the translation parameter uniquely, even when the differential equation and far value leave it free.

Using distinct fixed tags $t_1,\dots ,t_n$, encode a tuple $(a_1,\dots ,a_n)$ by the set $\{⟨t_i,a_i⟩:1\le i\le n\}$. Injectivity of the ordered-pair construction makes the unique member with first coordinate $t_i$ recover $a_i$. Thus a set can retain the order of a finite tuple when its entries are tagged, even if some coordinates coincide.

An ordered triple is a finite tuple with three ordered coordinates. Repeated coordinates are allowed; equality requires equality at each of the three positions.

For a nonempty finite sample, assign each distinct observed value its relative frequency. The observed values form the finite positive-mass support; the counts sum to the sample size, so the probabilities sum to one. Repeated observations increase the same outcome's mass rather than adding repeated support entries.

Given two finite probability distributions, assign the product of their probability masses to each ordered pair. The support is the Cartesian product of the two positive-mass supports. Distributivity gives ${\sum }_{x,y}{p}_1(x)p_2(y)=({\sum }_x{p}_1(x))({\sum }_y{p}_2(y))=1$. This is the joint distribution for independent coordinate outcomes; it is a construction from the marginals, not a claim that arbitrary jointly observed coordinates are independent.

Induct on the order of the finite p-group. Its center of a group is nontrivial unless the group itself is trivial. Quotient by the center to obtain a smaller finite p-group and lift a central series supplied by induction. Prepending the center gives a central series for the original group: the first term is central, and the remaining commutator-containment conditions are exactly the lifted quotient conditions. Hence every finite p-group is a nilpotent group. Since central factors are Abelian, every nilpotent group is also a soluble group.

Let a proposer divide a pie of size $M>0$ by continuously offering its responder $q\in [0,M]$. Rejection gives fixed continuation payoffs $b$ to the proposer and $c$ to the responder, with $0\le c<M$ and $b<M-c$. Then a subgame perfect equilibrium must have the responder accept $q=c$ with probability one. The proposer offers exactly $c$, retaining $M-c$.

Sequential rationality forces rejection for $q<c$ and acceptance for $q>c$. If acceptance at $c$ had probability $a<1$, that offer would give the proposer $a(M-c)+(1-a)b<M-c$. Every larger offer gives less than $M-c$ and can be improved by a smaller offer still above $c$; every smaller offer gives $b<M-c$. The supremum $M-c$ would therefore be unattained. Randomizing offers cannot attain it either: the nonnegative gap from $M-c$ would be strictly positive almost surely and hence have positive expectation. No best response would exist, contradicting equilibrium. Acceptance at $c$ removes this gap and makes $q=c$ the unique optimal proposal. The assumptions $c<M$ and $b<M-c$ exclude degenerate indifference of the proposer.

In finite-horizon alternating-offer bargaining, the proposer with $k$ rounds left retains fraction $P_k$ of the currently available pie. Terminal acceptance gives $P_1=1$. After rejection, the responder becomes proposer and can obtain fraction $\alpha P_{k-1}$ of the current pie. The continuous-offer bargaining acceptance lemma therefore gives $P_k=1-\alpha P_{k-1}$. Summing the resulting finite geometric series proves the displayed formula.

Odd horizons give shares above $1/(1+\alpha )$, decreasing toward it; even horizons give shares below it, increasing toward it. Between two available horizons $k_1<k_2$, choose $k_1$ if $k_1$ is odd and $k_2$ if $k_1$ is even. This concerns the initial proposer and immediate agreement, rather than waiting until the selected final round to obtain its share.

For $G\in C^1([0,1])$, the displayed expression solves $Tf=G$, where $T=H_{(0,1)}/i$ and $H_{(0,1)}$ is the finite Hilbert transform with denominator $t-x$. To derive it, put $S(z)=\sqrt{z(z-1)}\sim z$ at infinity, take $h$ to be the Cauchy transform of $i\rho G$, and set $w=h/S$. The Sokhotski–Plemelj formula gives $w_{+}+w_{-}=G$ and the displayed jump $f=w_{+}-w_{-}$. The endpoint behavior is at most inverse square root, and $w=O(z^{-2})$ at infinity. Cauchy reconstruction from a jump across a finite interval then gives $w=C[f]$ and hence $Tf=G$. Adding any constant multiple of $1/\rho$ gives another solution in the integrable endpoint class.

Multiplying an auxiliary sign-changing factor by a growing analytic factor can invalidate this reconstruction. For example, using $S$ instead of $1/S$ with $G=1$ produces the constant function $w=1/2$ and zero jump. The boundary sum alone is not enough: the entire remainder in reconstructing $w$ from its jump must also be controlled.

Triangulate the region $0\le x,y\le 2$, $x+y\ge 2$ by $x=1$, $y=1$, and $x+y=3$. Impose zero Dirichlet data on the horizontal and vertical exterior edges and zero Neumann data on the sloping edge. Only the midpoint $B=(1,1)$ of the sloping edge remains free in the continuous piecewise linear space. For source one, its stiffness entry is $1/2+1/2+1=2$, while its load is three triangle contributions of $1/6$. The single free value is therefore $1/4$.

A bilinear square finite element uses polynomials of degree at most one separately in each coordinate and takes values at the four corners as degrees of freedom. On the unit square its nodal shape functions are $(1-\xi )(1-\eta )$, $\xi (1-\eta )$, $\xi \eta$, and $(1-\xi )\eta$, in anticlockwise corner order. Each is a tensor product of one-dimensional linear shape functions. A function in this space need not be affine, since $\xi \eta$ has total degree two. On a physical square of side $h$, its Laplacian stiffness matrix has diagonal entries $2/3$, adjacent-corner entries $-1/6$ and opposite-corner entries $-1/3$. Its Robin boundary condition edge matrix has the two-node block $(h/6)\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ for a unit boundary coefficient.