Alphabeta Math
CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26
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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.

xk=1+(1)kx_k = 1 + (-1)^k, yk=1+(1)k+1y_k = 1 + (-1)^{k+1} give lim sup(xkyk)=0<4\limsup(x_k y_k) = 0 < 4

Statement refuted

That the submultiplicativity of For bounded nonnegative sequences, lim sup(xkyk)(lim supxk)(lim supyk)\limsup(x_k y_k) \le (\limsup x_k)(\limsup y_k) can be improved to an equality: that for all bounded nonnegative sequences (xk)(x_k), (yk)(y_k) of reals, lim supk(xkyk)=(lim supkxk)(lim supkyk).\limsup_{k}(x_k y_k) = \Big(\limsup_{k} x_k\Big)\Big(\limsup_{k} y_k\Big).

Facts & Assumptions

[L3]

Absolute value: t=1|t| = 1 forces t=1t = 1 or t=1t = -1 (Basic properties of the absolute value, Absolute value in an ordered field).

[L4]

Order and field arithmetic: 0<10 < 1, so 0<2=1+10 < 2 = 1 + 1 and 0<4=220 < 4 = 2 \cdot 2; 1+1=21 + 1 = 2, 11=01 - 1 = 0, 1+(1)=01 + (-1) = 0 and 1(1)=21 - (-1) = 2; a product with a zero factor is 00 (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property)).

[L5]

Submultiplicativity: for bounded nonnegative sequences, lim supk(xkyk)(lim supkxk)(lim supkyk)\limsup_k(x_k y_k) \le (\limsup_k x_k)(\limsup_k y_k), all three quantities being real (For bounded nonnegative sequences, lim sup(xkyk)(lim supxk)(lim supyk)\limsup(x_k y_k) \le (\limsup x_k)(\limsup y_k)).

Counterexample

technique · direct
1.1

Each sks_k is 11 or 1-1. When sk=1s_k = 1 the pair (xk,yk)(x_k, y_k) is (2,0)(2, 0), and when sk=1s_k = -1 it is (0,2)(0, 2). In either case 0xk20 \le x_k \le 2 and 0yk20 \le y_k \le 2, so both sequences are bounded and nonnegative, and xkyk=0x_k y_k = 0 because one of the two factors is 00.

givenL1L3L4
1.2

For every nn both cases occur at an index n\ge n: sen=1s_{e_n} = 1 with enne_n \ge n and son=1s_{o_n} = -1 with onno_n \ge n.

givenL1
2.1

Hence Tn(x)={0,2}T_n(x) = \{0, 2\} and Tn(y)={0,2}T_n(y) = \{0, 2\} for every nn, each with least upper bound 22 in R\overline{\mathbb{R}}, since 22 bounds both elements and belongs to the set; so lim supkxk=lim supkyk=2\limsup_k x_k = \limsup_k y_k = 2. The product sequence is constantly 00, so Tn(xy)={0}T_n(x y) = \{0\} and lim supk(xkyk)=0\limsup_k (x_k y_k) = 0.

step 1.1step 1.2L2L4
3.1

The hypotheses of [L5] are met by step 1.1, and the inequality it gives reads 022=40 \le 2 \cdot 2 = 4. Since 0<40 < 4, it is strict, so the equality asserted above fails for this pair and the refuted claim is false.

step 2.1step 1.1L4L5

Remarks

Depends on

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 70 results over 20 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources