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

ri[key] = ri[key0], key-cons(key, fs[ri[key0]]), ri[key0] ∈ fs, key0 ∈ ri, key ∈ ri, key0 ∈ ri → ri[key0] ∈ fs ∧ key-cons(key0, fs[ri[key0]]) ⊦ key = key0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information