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Proof step 5 from proof ri-cons-drop (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 6:
as ∩ range ri = ∅, key ∈ ri, ri[key] ∈ fs, key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ key-cons(key, (fs \ as)[ri[key]])

Rule argument

pos: left 4 terms: key abort: F discard: T computed: T

Additional information

Used terms: key