LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-25

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.

Multiplicative cancellation: $c \neq 0$ and $a c = b c \Rightarrow a = b$

Statement

In any field $F$ (Field), if $a, b, c \in F$ with $c \neq 0$ and $a c = b c$ , then $a = b$ .

Facts & Assumptions

Given: A field $F$ and elements $a, b, c \in F$ with $c \neq 0$ and $a c = b c$ .

[L1]

Field axioms (Field): since $c \neq 0$ , the inverse $c^{- 1}$ exists with $c \cdot c^{- 1} = 1$ ; multiplication is associative; and $1$ is the multiplicative identity, so $x \cdot 1 = x$ .

Proof

technique · direct

1.1

Since $c \neq 0$ , the inverse $c^{- 1}$ exists with $c \cdot c^{- 1} = 1$ .

1.2

We are given $a c = b c$ .

given

2.1

Using $c \cdot c^{- 1} = 1$ and associativity, $a = a \cdot 1 = a (c \cdot c^{- 1}) = (a c) c^{- 1}$ .

step 1.1L1

2.2

Using $c \cdot c^{- 1} = 1$ and associativity, $b = b \cdot 1 = b (c \cdot c^{- 1}) = (b c) c^{- 1}$ .

step 1.1L1

2.3

Since $a c = b c$ , we have $(a c) c^{- 1} = (b c) c^{- 1}$ .

step 1.2

3.1

Combining, $a = (a c) c^{- 1} = (b c) c^{- 1} = b$ .

step 2.1step 2.2step 2.3∎

Depends on

Field

Used by

Nothing in the library uses this result yet.

Dependency tree · one level

1 result within one dependency step of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 1 (standard reference, not scraped)
M. Spivak, Calculus, 4th ed., Ch. 1 (standard reference, not scraped)
Tom Denton, Introduction to Algebraic Structures: The Problem of Division (standard reference, not scraped)

Multiplicative cancellation: c≠0c=0 and ac=bc⇒a=bac=bc⇒a=b