Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13
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.

Normalization, nonnegativity, monotonicity, complements, and differences in a finite probability space

Statement

In a finite probability space (Ω,w), events ABΩ satisfy 0P(A)P(B)1, and P()=0,P(Ω)=1,P(Ac)=1P(A),P(BA)=P(B)P(A). Probability zero need not imply that an event is empty.

Facts & Assumptions

Given: A finite probability space (Ω,w) and events AB.

[L1]

Event probability is the sum of the nonnegative weights of its outcomes, and the sum of all outcome weights is 1 (Finite probability spaces, outcome weights, events, and event probabilities).

[L2]

Finite sums preserve nonnegativity and order (Laws of finite sums and finite products).

[L3]

The real numbers form a totally ordered field (The reals form a totally ordered field).

Proof

technique · direct
1.1

The empty sum is 0, while the sum over Ω is 1, so P()=0 and P(Ω)=1.

L1
1.2

Since every summand in P(A) is nonnegative, 0P(A). Splitting the sum over B into A and BA gives P(B)=P(A)+P(BA), so P(A)P(B).

L1L2
2.1

Taking B=Ω in step 1.2 gives P(Ac)=1P(A); rearranging the same identity for general AB gives P(BA)=P(B)P(A).

step 1.1step 1.2L3algebra
3.1

The definition permits zero weights, so a singleton outcome of weight zero is a nonempty event of probability zero. All displayed conclusions follow.

L1step 1.1step 1.2step 2.1

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 52 results over 17 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