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Proof step 2 from proof abs-keys-orphans2 (previous step)

abs-keys(orphans(fs0, ri0)) = abs-keys(orphans(fs, ri)) ⊦ abs-keys(orphans(fs0, ri0)) = abs-keys(orphans(fs, ri)) ↔ orphans(fs0, ri0) = orphans(fs, ri)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information