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Proof step 4 from proof ri-cons-key (previous step)

key ∈ ri, key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]), ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ fs[ri[key]].key = key

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information