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Proof step 3 from proof ri-cons-drop (previous step)

as ∩ range ri = ∅, key ∈ ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ ri[key] ∈ fs \ as ∧ key-cons(key, (fs \ as)[ri[key]])

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 4:
as ∩ range ri = ∅, key ∈ ri, ri[key] ∈ fs, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ ri[key] ∈ fs \ as ∧ key-cons(key, (fs \ as)[ri[key]])
Proof step 14:
as ∩ range ri = ∅, key ∈ ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ ri[key] ∈ fs, ri[key] ∈ fs \ as ∧ key-cons(key, (fs \ as)[ri[key]])

Rule argument

fma: ri[key] ∈ fs

Additional information

Used formulas: ri[key] ∈ fs