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Proof step 14 from proof ri-cons-drop (previous step)

as ∩ range ri = ∅, key ∈ ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ ri[key] ∈ fs, ri[key] ∈ fs \ as ∧ key-cons(key, (fs \ as)[ri[key]])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
as ∩ range ri = ∅, key ∈ ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ ri[key] ∈ fs, ri[key] ∈ fs \ as

Rule argument

Additional information

Used simplifier rules: ⊦ adr ∈ fs \ as ↔ adr ∈ fs ∧ ¬ adr ∈ as; (ws)