LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03

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.

In a module, $0_{R} m = 0_{M}$ , $r 0_{M} = 0_{M}$ , $(- r) m = - (r m)$ and $r (- m) = - (r m)$

For every left $R$ -module $M$ , scalar $r \in R$ , and element $m \in M$ ,

$0_{R} m = 0_{M}, r 0_{M} = 0_{M}, (- r) m = - (r m), r (- m) = - (r m) .$

Given: A left $R$ -module $M$ , $r \in R$ , and $m \in M$ .

[L1]

The module action distributes over both addition operations, and $(M, +, 0_{M})$ is an abelian group (Unital left and right modules over a ring; unqualified module means left module).

[L2]

[L3]

technique · direct

1.1L1L3given

Since $0_{R} = 0_{R} + 0_{R}$ , distributivity gives $0_{R} m = 0_{R} m + 0_{R} m$ ; cancellation yields $0_{R} m = 0_{M}$ .

1.2L1L3given

Since $0_{M} = 0_{M} + 0_{M}$ , distributivity gives $r 0_{M} = r 0_{M} + r 0_{M}$ ; cancellation yields $r 0_{M} = 0_{M}$ .

2.1step 1.1L1L2L3given

From $(r + (- r)) m = 0_{R} m = 0_{M}$ , distributivity and step 1.1 give $r m + (- r) m = 0_{M}$ , so $(- r) m = - (r m)$ .

3.1step 1.2L1L3given∎

From $r (m + (- m)) = r 0_{M} = 0_{M}$ , distributivity and step 1.2 give $r m + r (- m) = 0_{M}$ , so $r (- m) = - (r m)$ .

Depends on