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Proof step 6 from proof ri-cons-drop (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information