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Proof step 17 from proof in-ri-update-in-ri (previous step)

# ax1 = # ax0, # keys(fs, ax1) = # ax1, adr ∈ ri, ax1.toSet ⊆ range ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]), key = fs[adr].key ∨ key ∈ keys(fs, ax1) ⊦ key ∈ keys(fs, ax1), key ∈ ri

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
# ax1 = # ax0, adr ∈ ri, ax1.toSet ⊆ range ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ fs[adr].key ∈ keys(fs, ax1), fs[adr].key ∈ ri

Rule argument

Additional information

Used simplifier rules: ⊦ # keys(fs, ax) = # ax; (s)