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

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 4:
key0 = key, key0 ∈ ri[key, adr], key ∈ ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ node-closed(fs[ri[key]], ri, fs, ro), ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ adr ∈ fs, (ri[key, adr][key0] = adr ∨ ri[key, adr][key0] ∈ fs) ∧ node-closed(fs[adr, fs[ri[key]]][ri[key, adr][key0]], ri[key, adr], fs[adr, fs[ri[key]]], ro)
Proof step 27:
key0 ∈ ri[key, adr], key ∈ ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ node-closed(fs[ri[key]], ri, fs, ro), ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ key0 = key, adr ∈ fs, (ri[key, adr][key0] = adr ∨ ri[key, adr][key0] ∈ fs) ∧ node-closed(fs[adr, fs[ri[key]]][ri[key, adr][key0]], ri[key, adr], fs[adr, fs[ri[key]]], ro)

Rule argument

fma: key0 = key

Additional information

Used formulas: key0 = key