ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)

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.

Worked reverse Cauchy-Schwarz and reverse triangle inequalities

Example

Worked computation in the chart $x$ of an inertial frame with $η_{0} = diag (- 1, 1, 1, 1)$ ; the vectors below are coordinate columns in the frame's basis.

(a) For $v : = (5, 3, 0, 0)$ and $w : = (5, - 3, 0, 0)$ one has $η (v, v) = η (w, w) = - 25 + 9 = - 16$ , so the timelike lengths are $\sqrt{16} = 4$ each, and $η (v, w) = - 25 - 9 = - 34$ , so $- η (v, w) = 34 \geq 4 \cdot 4 = 16$ ; the inequality is strict because $v$ and $w$ are not positive proportional.

(b) For $w^{'} : = 2 v = (10, 6, 0, 0)$ one has $η (w^{'}, w^{'}) = - 64$ and $\sqrt{- η (w^{'}, w^{'})} = 8$ , while $η (v, w^{'}) = - 50 + 18 = - 32$ , so $- η (v, w^{'}) = 32 = 4 \cdot 8$ : this is the equality case of the reversed Cauchy-Schwarz inequality, attained by the positive proportional pair $(v, 2 v)$ .

(c) Reverse triangle: $v + w = (10, 0, 0, 0)$ has $- η (v + w, v + w) = 100$ , so its root is $10 \geq 4 + 4 = 8$ , again strict because $v$ and $w$ are not positive proportional. For $u : = (5, 0, 0, 0)$ , of timelike length $5$ , one has $- η (v, u) = 25 \geq 4 \cdot 5 = 20$ and $v + u = (10, 3, 0, 0)$ with $\sqrt{- η (v + u, v + u)} = \sqrt{100 - 9} = \sqrt{91} \geq 4 + 5 = 9$ , since $91 > 81$ .

(d) Each of $v$ , $w$ , $w^{'}$ and $u$ is future timelike ( $5 > 3$ , $5 > 3$ , $10 > 6$ and $5 > 0$ in the first coordinate against the spatial norm), so all inequalities above are instances of the future timelike hypotheses of The reversed Cauchy-Schwarz inequality for future timelike vectors and The reversed triangle inequality and convexity of the future timelike cone.

Facts & Assumptions

Given: An inertial frame with chart $x$ and $η_{0} = diag (- 1, 1, 1, 1)$ , and the coordinate vectors $v = (5, 3, 0, 0)$ , $w = (5, - 3, 0, 0)$ , $w^{'} = (10, 6, 0, 0)$ , $u = (5, 0, 0, 0)$ (Inertial frames and inertial charts of the affine Minkowski model).

[L1]

Pairing in the frame: $η (r, r^{'}) = - r^{0} r^{' 0} + r^{1} r^{' 1} + r^{2} r^{' 2} + r^{3} r^{' 3}$ for coordinate vectors $r, r^{'}$ (Inertial frames and inertial charts of the affine Minkowski model, Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space).

[L2]

Causal classification and cone coordinates: $r$ is timelike when $η (r, r) < 0$ , and in a frame with future first vector $r$ is future timelike exactly when $r^{0} > ∣ r_{s} ∣$ with $∣ r_{s} ∣ = \sqrt{(r^{1})^{2} + (r^{2})^{2} + (r^{3})^{2}}$ (Timelike, null, spacelike and causal vectors of a Lorentz vector space, The timelike set has exactly two convex components, described in orthonormal coordinates).

[L3]

Reversed Cauchy-Schwarz inequality: for future timelike $r, s$ with $a = \sqrt{- η (r, r)}$ , $b = \sqrt{- η (s, s)}$ one has $- η (r, s) \geq a b$ , with equality exactly when $r$ and $s$ are positive proportional (The reversed Cauchy-Schwarz inequality for future timelike vectors).

[L4]

Reversed triangle inequality: for future timelike $r, s$ one has $r + s$ future timelike and $\sqrt{- η (r + s, r + s)} \geq \sqrt{- η (r, r)} + \sqrt{- η (s, s)}$ , with equality exactly when $r$ and $s$ are positive proportional (The reversed triangle inequality and convexity of the future timelike cone).

[L5]

Square roots and monotonicity: every real $c \geq 0$ has a unique $\sqrt{c} \geq 0$ with $(\sqrt{c})^{2} = c$ , so in particular $\sqrt{16} = 4$ ; for $x, y \geq 0$ one has $x < y ⟺ x^{2} < y^{2}$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Squaring is monotone on the nonnegatives).

[L6]

Elementary order arithmetic in $R$ : the order is trichotomous and transitive (Ordered field).

Verification

technique · direct

1.1givenL1algebra

Evaluating the diagonal form [L1] on the given coordinate vectors gives $η (v, v) = - 25 + 9 = - 16$ , $η (w, w) = - 25 + 9 = - 16$ , $η (w^{'}, w^{'}) = - 100 + 36 = - 64$ , $η (u, u) = - 25$ , $η (v, w) = - 25 - 9 = - 34$ , $η (v, w^{'}) = - 50 + 18 = - 32$ , $η (v, u) = - 25$ , $η (v + w, v + w) = η ((10, 0, 0, 0), (10, 0, 0, 0)) = - 100$ and $η (v + u, v + u) = - 100 + 9 = - 91$ .

2.1step 1.1L2algebra

By [L2] each of the four vectors is future timelike: $v^{0} = 5 > 3 = ∣ v_{s} ∣$ , $w^{0} = 5 > 3 = ∣ w_{s} ∣$ , $(w^{'})^{0} = 10 > 6 = ∣ (w^{'})_{s} ∣$ and $u^{0} = 5 > 0 = ∣ u_{s} ∣$ ; this is (d).

3.1step 1.1step 2.1L3L5L6algebra

For (a): $\sqrt{- η (v, v)} = \sqrt{- η (w, w)} = \sqrt{16} = 4$ by [L5], so the right-hand side is $16$ , while $- η (v, w) = 34$ ; the pair is not positive proportional, since $w = t v$ would give $t = 1$ from the first coordinates and then $- 3 = 3 t$ , a contradiction, and $v = t w$ fails the same way; hence the strict clause of [L3] gives $34 > 16$ , as claimed.

3.2step 1.1step 2.1L3L5algebra

For (b): $w^{'} = 2 v$ with $2 > 0$ , so the pair $(v, w^{'})$ is positive proportional; by the equality case of [L3] applied to the future timelike pair $(v, w^{'})$ one gets $- η (v, w^{'}) = \sqrt{- η (v, v)} \sqrt{- η (w^{'}, w^{'})}$ , and the values of step 1.1 give $\sqrt{- η (w^{'}, w^{'})} = \sqrt{64} = 8$ and $32 = 4 \cdot 8$ , the displayed equality.

4.1step 1.1step 2.1step 3.1L3L4L5algebra

For (c): the sum $v + w$ has the coordinate column $(10, 0, 0, 0)$ , so $- η (v + w, v + w) = 100$ and $\sqrt{100} = 10$ by [L5], and [L4] gives $10 \geq 4 + 4 = 8$ ; equality would force $v$ and $w$ positive proportional by [L4], which step 3.1 excludes, so the inequality is strict. Similarly $- η (v, u) = 25 \geq 4 \cdot 5 = 20$ with strictness by [L3] since $v$ and $u$ are not positive proportional, and $v + u$ has coordinate column $(10, 3, 0, 0)$ with $- η (v + u, v + u) = 91$ ; since $91 > 81 = 9^{2}$ with $9 \geq 0$ and $\sqrt{91} \geq 0$ , monotonicity of squaring on the nonnegatives [L5] gives $\sqrt{91} > 9 = 4 + 5$ .

5.1step 2.1step 3.1step 3.2step 4.1L3L4∎

Steps 3.1, 3.2 and 4.1 verify the strict, equality and strict reverse-triangle instances of (a), (b) and (c) respectively, all under the future timelike hypotheses confirmed in step 2.1; the equality case occurs exactly at the positive proportional pair of (b), in agreement with [L3] and [L4].

Worked reverse Cauchy-Schwarz and reverse triangle inequalities

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources