FA-1787 / Relational algebra / Member archive

Semi join: Uses absence instead of existence · case 02

The query violates its result contract: uses absence instead of existence.

Member previewVariant 2 · 3 implementations · 8 checks per implementation

Case contract

Return each a ID once if any b has the same known k. Inputs are finite relations a=rows and b=refs with columns (id,k,v,w,flag,parent). IDs are unique within each relation, None is SQL NULL, keys are ASCII strings or None, and other values are integers or None. Return an unordered result bag as lists sorted by repr, preserving multiplicity unless explicitly distinct.

Why this case matters

Executes actual in-memory SQLite relational semantics. The fixtures exercise empty inputs, nullable groups, duplicate facts, known zero, negative values, peer ties and partition boundaries. This models the specified query contract, not optimizer performance, disk durability or a complete migration transaction.

One recorded failure

Sample boundary fixture

This sample comes from the broken implementation of a controlled reproducer.

Boundary fixture	Actual	Expected	Outcome
left facts without references	`[[102], [112], [122], [12], [132], [142], [152], [162], [172], [182], [192], [202], [22], [32], [42], [52], [62], [72], [82], [92]]`	`[]`	Failed

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