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Proof step 2 from proof key-in-ri-in-fs (
previous step
)
ri[key] = adr, ri-cons(ri, fs), key ∈ ri ⊦ adr ∈ fs
Heuristic simplifier applied rule simplifier
And closed the goal!
Rule argument
Additional information
Used simplifier rules: ⊦ key ∈ ri ∧ ri-cons(ri, fs) → ri[key] ∈ fs; (ss)