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Proof step 2 from proof in-ri-update-in-ri (previous step)

# ax0 = # ax, ri-cons(ri, fs), ax0.toSet ⊆ range ri ⊦ key ∈ ri[keys(fs, ax0), ax] ↔ key ∈ ri

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# ax0 = # ax, ax0.toSet ⊆ range ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ key ∈ ri[keys(fs, ax0), ax] ↔ key ∈ ri

Rule argument

Additional information

Used simplifier rules: ri-cons: ⊦ ri-cons(ri, fs) ↔ (∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]));