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Proof step 10 from proof ri-cons-gc-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
key ∈ ri, key0 ∈ ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ ri[key0] = adr, key0 = key, adr ∈ fs, ri[key0] ∈ fs ∧ key-cons(key0, fs[ri[key0]])

Rule argument

Additional information

Used simplifier rules: ⊦ adr ∈ fs ∧ ¬ adr0 ∈ fs → fs[adr0, nd][adr] = fs[adr]; (s)