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

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

User applied rule all left

And obtained the following new goals

Proof step 4:
ri[key] = ri[key0], key0 ∈ ri, key ∈ ri, key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]), ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ key = key0

Rule argument

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

Additional information

Used terms: key