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Proof step 6 from proof range-in-ri-keys-in-ri (previous step)

ri-cons(ri, fs), adr ∈ ri, ax0.toSet ⊆ range ri, ri-cons(ri, fs) ∧ ax0.toSet ⊆ range ri → keys(fs, ax0).toSet ⊆ dom ri ⊦ keys(fs, ax0).toSet ++ fs[adr].key ⊆ dom ri

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
ri-cons(ri, fs), adr ∈ ri, ax0.toSet ⊆ range ri, keys(fs, ax0).toSet ⊆ dom ri ⊦ fs[adr].key ∈ ri

Rule argument

Additional information

Used simplifier rules: ⊦ key ∈ dom ri ↔ key ∈ ri; (s)% ⊦ ks ++ key ⊆ ks0 ↔ ks ⊆ ks0 ∧ key ∈ ks0; (ws)