LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12

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.

Quotient operations are well defined

Statement

For every ideal $I$ in a Boolean algebra $B$ , the relation $\sim_{I}$ is an equivalence and a congruence. The stated operations make $B / I$ Boolean. The quotient map $q (a) = [a]$ is a surjective homomorphism with kernel $I$ . For every Boolean homomorphism $h : B \to C$ with $h [I] = {0}$ there is exactly one Boolean homomorphism $\overset{ˉ}{h} : B / I \to C$ satisfying $\overset{ˉ}{h} q = h$ . All assertions hold in ZF, including $I = B$ .

Facts & Assumptions

[F1]

Boolean homomorphisms and quotient relation defines $△$ , $\sim_{I}$ , the class operations and bound-preserving homomorphisms.

Proof

Given: A Boolean algebra $B$ , an ideal $I \subseteq B$ , and the definitions in F1.

1.1F1algebra

For $a, b, c \in B$ , split $a \land \neg c$ by $b \lor \neg b = 1$ . Its two pieces lie respectively below $b \land \neg c$ and $a \land \neg b$ . Split $c \land \neg a$ the same way. Joining these four inequalities gives $a △ c \leq (a △ b) \lor (b △ c)$ . Also $a △ a = 0$ and $a △ b = b △ a$ . Since an ideal contains $0$ , is downward closed and is closed under joins, these identities prove reflexivity, symmetry and transitivity of $\sim_{I}$ .

1.2F1algebra

For $a, b, c, d \in B$ , expand $(a \land c) \land \neg (b \land d)$ using $\neg (b \land d) = \neg b \lor \neg d$ . The resulting pieces lie below $a \land \neg b$ and $c \land \neg d$ . The reverse difference has pieces below $b \land \neg a$ and $d \land \neg c$ . Hence $(a \land c) △ (b \land d) \leq (a △ b) \lor (c △ d)$ . Moreover $(\neg a) △ (\neg b) = a △ b$ . Thus equivalent inputs give equivalent meets and complements; the identity $a \lor c = \neg (\neg a \land \neg c)$ gives the same for joins.

2.1F1step 1.1step 1.2algebra

Steps 1.1 and 1.2 make every proposed operation a function on classes. Each Boolean identity in the quotient is the class of that identity in $B$ , so the quotient is Boolean. The map $q$ preserves each operation by its definition and is onto because every class has the form $[a]$ . Finally $[a] = [0]$ if and only if $a △ 0 = a \in I$ , proving the kernel assertion. If $I = B$ , every pair is equivalent and the quotient has one element.

3.1F1step 2.1algebra

Let $h$ be as in the statement. If $a \sim_{I} b$ , preservation of the operations gives $h (a) △ h (b) = h (a △ b) = 0$ . For any Boolean elements $u, v$ , $u △ v = 0$ implies $u \land \neg v = v \land \neg u = 0$ , hence $u \leq v \leq u$ and $u = v$ . Therefore $\overset{ˉ}{h} ([a]) = h (a)$ is well defined. Substituting this formula into each quotient operation proves preservation of both bounds, meet, join and complement, and $\overset{ˉ}{h} q = h$ . Any factorization must take $[a]$ to $h (a)$ because $q$ is onto, proving uniqueness without a choice of representatives. QED.

Depends on

Boolean homomorphisms and quotient relation

Dependency tree · 2 levels

Sources

Tressl, Stone Duality for Boolean Algebras, 2.1.1 and 2.3.1 (Boolean laws); local quotient calculation