Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-06 (claude-opus-5)
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.

If aba \subseteq b then ab\bigcup a \subseteq \bigcup b; if in addition aa \neq \varnothing then ba\bigcap b \subseteq \bigcap a; and cac \subseteq \bigcup a for every cac \in a, while ac\bigcap a \subseteq c for every cac \in a

Statement

Let aa and bb be sets.

  • (i) If aba \subseteq b then ab\bigcup a \subseteq \bigcup b.
  • (ii) If aba \subseteq b and aa \neq \varnothing, then bb \neq \varnothing, both intersections are defined, and ba\bigcap b \subseteq \bigcap a.
  • (iii) cac \subseteq \bigcup a for every cac \in a.
  • (iv) ac\bigcap a \subseteq c for every cac \in a.

Facts & Assumptions

Given: sets aa and bb.

[L1]

x\bigcup x is the set whose elements are exactly the elements of the elements of xx (The union x\bigcup x of a set, and the binary union ab:={a,b}a \cup b := \bigcup \{a,b\}).

[L2]

For xx \neq \varnothing, x\bigcap x is the set whose elements are exactly the sets belonging to every element of xx (The intersection x\bigcap x of a nonempty set, the binary intersection ab:={a,b}a \cap b := \bigcap\{a,b\}, and disjointness).

[L4]

There is exactly one set with no elements, written \varnothing (There is exactly one set with no elements, written \varnothing).

Proof

technique · direct
1.1

Claim (i): assume aba \subseteq b and let zaz \in \bigcup a; then zsz \in s for some sas \in a, and sbs \in b because aba \subseteq b, so zbz \in \bigcup b.

L1L3
1.2

Claim (ii): assume aba \subseteq b and aa \neq \varnothing. Then aa has a member, which is also a member of bb, so bb \neq \varnothing and both intersections are defined. If zbz \in \bigcap b then zz lies in every member of bb, hence in every member of aa, so zaz \in \bigcap a.

L2L3L4
1.3

Claim (iii): let cac \in a and zcz \in c; then zz lies in a member of aa, so zaz \in \bigcup a, and therefore cac \subseteq \bigcup a.

L1L3
1.4

Claim (iv): let cac \in a; then aa \neq \varnothing, so a\bigcap a is defined, and every zaz \in \bigcap a lies in every member of aa, in particular in cc; therefore ac\bigcap a \subseteq c.

L2L3L4
2.1

Claims (i) to (iv) are established, which is the statement.

step 1.1step 1.2step 1.3step 1.4

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: 11 results over 5 levels. 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