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Proof step 2 from proof ri-cons-key (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
key ∈ ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ fs[ri[key]].key = key

Rule argument

Additional information

Used simplifier rules: ri-cons: ⊦ ri-cons(ri, fs) ↔ (∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]));