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Proof step 4 from proof ri-cons-gc-one (previous step)

key0 = key, key0 ∈ ri[key, adr], key ∈ ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ adr ∈ fs, (ri[key, adr][key0] = adr ∨ ri[key, adr][key0] ∈ fs) ∧ key-cons(key0, fs[adr, fs[ri[key]]][ri[key, adr][key0]])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
key ∈ ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ key-cons(key, fs[ri[key]]), adr ∈ fs

Rule argument

Additional information

Used simplifier rules: ⊦ fs[adr, nd][adr] = nd; (s)% ⊦ ri[key, adr][key] = adr; (s)% ⊦ key ∈ ri[key, adr]; (s)