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Proof step 13 from proof ri-cons-drop (previous step)

key-cons(key, fs[ri[key]]), ri[key] ∈ as, key ∈ ri, ri[key] ∈ fs, ∀ adr. ¬ adr ∈ as ∩ range ri ⊦ # as = 0, as = ∅, key-cons(key, (∅)[ri[key]])

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: ri[key]

Additional information

Used simplifier rules: ⊦ adr ∈ as ∩ as0 ↔ adr ∈ as ∧ adr ∈ as0; (ws)% ⊦ adr ∈ range ri ↔ adr ∈ ri; (s)%key ∈ ri ⊦ ri[key] ∈ ri; (s)%int-a: ⊦ (as ∩ as0) ∩ as1 = as ∩ as0 ∩ as1; from specification set-union[node-store]%int-c: ⊦ as ∩ as0 = as0 ∩ as; from specification set-union[node-store] Used terms: ri[key]