TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30

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.

The product construction gives a DFA for language difference

Statement

Let $D_{1}$ and $D_{2}$ be DFA's over the same alphabet $Σ$ , with recognized languages $L_{1}$ and $L_{2}$ . Then there is a DFA over $Σ$ recognizing $L_{1} ∖ L_{2} .$

Facts & Assumptions

Given: Two DFA's $D_{1} = (Q_{1}, Σ, δ_{1}, q_{1}, F_{1})$ and $D_{2} = (Q_{2}, Σ, δ_{2}, q_{2}, F_{2})$ over the same alphabet.

[L1]

For languages over a fixed alphabet, the difference is $L ∖ K$ , by Boolean operations on languages over a fixed alphabet.

[L2]

The product construction tracks the two component runs in parallel, by The product construction gives DFA's for union and intersection.

[L3]

A word is accepted exactly when its final state lies in the accepting set, by Acceptance of a word by a DFA and the recognized language.

Proof

technique · direct

1.1

Use the same product machine as in [L2], with state set $Q_{1} \times Q_{2}$ , start state $(q_{1}, q_{2})$ , and product transition function. This machine reaches the pair of component states reached by $D_{1}$ and $D_{2}$ on the same input word.

L2construct

1.2

Declare a pair $(p, r)$ accepting exactly when $p \in F_{1}$ and $r \notin F_{2}$ , that is, take accepting set $F : = F_{1} \times (Q_{2} ∖ F_{2})$ .

construct

2.1

By step 1.1 and [L3], a word $w$ is accepted by this product machine exactly when $D_{1}$ accepts $w$ and $D_{2}$ does not accept $w$ .

L3step 1.1step 1.2

3.1

By [L1], that condition is exactly $w \in L_{1} ∖ L_{2}$ . So the constructed machine recognizes the language difference.

L1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 results within two dependency steps 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

Jean Gallier and Jocelyn Quaintance, Introduction to the Theory of Computation: Some Notes for CIS511 (standard reference, not scraped)