ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Parallel future null displacements compose to a future null displacement

Example

Worked computation in the chart $x$ of an inertial frame (Inertial frames and inertial charts of the affine Minkowski model). Let $v : = (1, 1, 0, 0)$ and $w : = (2, 2, 0, 0)$ be coordinate columns with respect to the basis of the frame.

(a) Both vectors are future null: $η (v, v) = - 1 + 1 = 0$ and $η (w, w) = - 4 + 4 = 0$ with positive time components, and $w = 2 v$ , so the two directions are parallel (Timelike, null, spacelike and causal vectors of a Lorentz vector space).

(b) The sum $v + w = (3, 3, 0, 0)$ is again future null, since $η (v + w, v + w) = - 9 + 9 = 0$ and its time component is positive, and the endpoint $q : = x^{- 1} (3, 3, 0, 0)$ lies in $J^{+} (p)$ for $p : = x^{- 1} (0, 0, 0, 0)$ but not in $I^{+} (p)$ , by the coordinate criterion (The closed future cone in inertial coordinates: description, finite sums, and the coordinate criterion for causal reachability, Causal reachability, causal futures and pasts, causal diamonds, and reachability by future causal paths in affine Minkowski space); the straight path $s \mapsto p + s (q - p)$ has constant tangent $v + w$ , so it is a straight future null path from $p$ to $q$ .

(c) This is the boundary case of the sum clause of the same lemma: for nonzero future causal vectors $u$ and $z$ the sum $u + z$ is future causal, and it is future null exactly when $u$ and $z$ are both future null and positive multiples of one another, as here. By contrast the explicit nonparallel future null pair $(1, 1, 0, 0)$ and $(1, - 1, 0, 0)$ has sum $(2, 0, 0, 0)$ , whose squared norm is $- 4$ : the sum of two future null displacements is null only in the parallel case.

Facts & Assumptions

Given: An inertial frame with chart $x$ and $η_{0} = diag (- 1, 1, 1, 1)$ , the vectors $v = (1, 1, 0, 0)$ and $w = (2, 2, 0, 0)$ , the events $p = x^{- 1} (0, 0, 0, 0)$ and $q = x^{- 1} (3, 3, 0, 0)$ , and the pair $u_{0} = (1, 1, 0, 0)$ , $z_{0} = (1, - 1, 0, 0)$ .

[L1]

Coordinate criterion and cone description: for a vector $v = (v^{0}, v_{s})$ with spatial norm $∣ v_{s} ∣$ , $v \in \overline{C^{+}}$ iff $v^{0} \geq ∣ v_{s} ∣$ , $v \in C^{+}$ iff $v^{0} > ∣ v_{s} ∣$ , and $v$ is future null exactly when $∣ v_{s} ∣ = v^{0} > 0$ ; for events $r, s$ with $D = x (s) - x (r)$ , $s \in J^{+} (r)$ iff $D^{0} \geq ∣ D_{s} ∣$ and $s \in I^{+} (r)$ iff $D^{0} > ∣ D_{s} ∣$ (The closed future cone in inertial coordinates: description, finite sums, and the coordinate criterion for causal reachability).

[L2]

Causal classification: $η (v, v) = - (v^{0})^{2} + ∣ v_{s} ∣^{2}$ in the inertial basis; $v$ is null exactly when $v \neq 0$ and $η (v, v) = 0$ , and timelike when $η (v, v) < 0$ (Timelike, null, spacelike and causal vectors of a Lorentz vector space); the sum of two elements of $\overline{C^{+}}$ lies in $\overline{C^{+}}$ , and if one of them lies in $C^{+}$ the sum lies in $C^{+}$ (The closed future cone in inertial coordinates: description, finite sums, and the coordinate criterion for causal reachability).

[L3]

Euclidean structure: for spatial coordinate vectors $p, q$ the norm satisfies $∣ p ∣^{2} = \sum_{i} (p^{i})^{2}$ , $∣ p ∣ = 0$ implies $p = 0$ , the triangle inequality $∣ p + q ∣ \leq ∣ p ∣ + ∣ q ∣$ holds, and $∣ p + q ∣^{2} = ∣ p ∣^{2} + 2 p \cdot q + ∣ q ∣^{2}$ with the Euclidean scalar product $p \cdot q$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , The Euclidean spatial structure and cross product of an inertial frame, The inner-product norm is definite, homogeneous, and satisfies the triangle inequality).

[L4]

Chart linearity: coordinate differences of events are the coordinates of their displacement (Inertial frames and inertial charts of the affine Minkowski model).

Verification

technique · direct

1.1givenL1L2algebra

The evaluations $η (v, v) = - 1 + 1 = 0$ , $η (w, w) = - 4 + 4 = 0$ and the displayed $w = 2 v$ hold by the diagonal form of [L2]; both vectors are nonzero with positive time components, so [L1] makes them future null; this is (a).

2.1givenstep 1.1L1L2L4algebra

The sum is $v + w = (3, 3, 0, 0)$ with $η (v + w, v + w) = - 9 + 9 = 0$ and time component $3 > 0$ , so it is future null by [L1]; by [L4] the displacement $q - p$ has coordinates $(3, 3, 0, 0)$ , so the criterion gives $3 \geq 3$ , that is $q \in J^{+} (p)$ , while $3 > 3$ is false, that is $q \notin I^{+} (p)$ . The straight path $s \mapsto p + s (q - p)$ has constant tangent $q - p$ with coordinates $v + w$ , a future null vector, so it is a straight future null path from $p$ to $q$ ; this is (b).

2.2step 1.1L1L2L3algebra

For the characterization let $u = (a, p)$ and $z = (b, q)$ be nonzero future causal vectors written in the inertial basis, so that $a \geq ∣ p ∣$ , $b \geq ∣ q ∣$ and $a, b > 0$ because a future causal vector with vanishing time component has vanishing spatial part. If $u + z$ is null then $(a + b)^{2} = ∣ p + q ∣^{2}$ , so by [L3] $a + b = ∣ p + q ∣ \leq ∣ p ∣ + ∣ q ∣ \leq a + b$ and every inequality is an equality; in particular $∣ p ∣ = a$ , $∣ q ∣ = b$ and $∣ p + q ∣ = ∣ p ∣ + ∣ q ∣$ . Expanding with [L3], $∣ p + q ∣^{2} = ∣ p ∣^{2} + 2 p \cdot q + ∣ q ∣^{2}$ equals $(∣ p ∣ + ∣ q ∣)^{2} = ∣ p ∣^{2} + 2 ∣ p ∣ ∣ q ∣ + ∣ q ∣^{2}$ , hence $p \cdot q = ∣ p ∣ ∣ q ∣$ ; since $a = ∣ p ∣ > 0$ , the vector $q - \frac{b}{a} p$ has squared norm $∣ q ∣^{2} - 2 \frac{b}{a} p \cdot q + \frac{b^{2}}{a^{2}} ∣ p ∣^{2} = b^{2} - 2 b^{2} + b^{2} = 0$ , so $q = \frac{b}{a} p$ by positive definiteness of the norm and $z = \frac{b}{a} u$ with $\frac{b}{a} > 0$ ; moreover $u$ and $z$ are future null because $∣ p ∣ = a$ and $∣ q ∣ = b$ with $a, b > 0$ . Conversely, if $z = λ u$ with $λ > 0$ and $u$ future null, then $u + z = (1 + λ) u$ has $η ((1 + λ) u, (1 + λ) u) = (1 + λ)^{2} η (u, u) = 0$ and positive time component, so it is future null; this proves the characterization of (c).

3.1step 2.1step 2.2L1L2algebra∎

For the nonparallel pair, $u_{0} = (1, 1, 0, 0)$ and $z_{0} = (1, - 1, 0, 0)$ are future null by [L1] since $∣ (1) ∣ = 1 = ∣ (1) ∣$ and $∣ (- 1) ∣ = 1$ with time component $1 > 0$ , while their sum is $(2, 0, 0, 0)$ with $η = - 4 < 0$ , a future timelike vector by [L2]: the sum of two future null displacements is therefore null exactly in the parallel case established in step 2.2, which is the boundary case of the sum clause illustrated by $v$ and $w$ in steps 1.1 and 2.1; all computations are exact integer evaluations and no choice principle is used.

Parallel future null displacements compose to a future null displacement

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources