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Proof step 3 from proof ri-cons-gc-one (previous step)

key ∈ ri, key0 ∈ ri[key, adr], ∀ 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 weak cut applied rule cut formula

And obtained the following new goals

Rule argument

fma: key0 = key

Additional information

Used formulas: key0 = key