Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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≠0c \ne 0 and ac=bc⇒a=bac = bc \Rightarrow a = b

Statement

In any field FF (Field), if a,b,c∈Fa, b, c \in F with c≠0c \ne 0 and ac=bcac = bc, then a=ba = b.

Facts & Assumptions

Given: A field FF and elements a,b,c∈Fa, b, c \in F with c≠0c \ne 0 and ac=bcac = bc.

[L1]

Field axioms (Field): since c≠0c \ne 0, the inverse c−1c^{-1} exists with c⋅c−1=1c \cdot c^{-1} = 1; multiplication is associative; and 11 is the multiplicative identity, so x⋅1=xx \cdot 1 = x.

Proof

technique · direct
1.1

Since c≠0c \ne 0, the inverse c−1c^{-1} exists with c⋅c−1=1c \cdot c^{-1} = 1.

L1
1.2

We are given ac=bcac = bc.

given
2.1

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

step 1.1L1
2.2

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

step 1.1L1
2.3

Since ac=bcac = bc, we have (ac)c−1=(bc)c−1(ac) c^{-1} = (bc) c^{-1}.

step 1.2
3.1

Combining, a=(ac)c−1=(bc)c−1=ba = (ac) c^{-1} = (bc) c^{-1} = b.

step 2.1step 2.2step 2.3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 1 result over 1 level. 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