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Proof step 8 from proof lookup-rem-inj-range-inter-empty (previous step)

ri-inj(ri), key ∈ ks, key ∈ ri, ∃ key0. key0 ∈ ri \ ks ∧ (ri \ ks)[key0] = ri[key] ⊦ # ks = 0, #(ri[ks] ∩ range(ri \ ks)) = 0, ks = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
ri[key0] = ri[key], ri-inj(ri), key ∈ ri, key ∈ ks, key0 ∈ ri ⊦ #(ri[ks] ∩ range(ri \ ks)) = 0, # ks = 0, ks = ∅, key0 ∈ ks

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ key ∈ ks → (ri \ ks)[key] = ri[key]; (s)% ⊦ key ∈ ri \ ks ↔ key ∈ ri ∧ ¬ key ∈ ks; (ws)