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Proof step 11 from proof ri-cons-gc-one (
previous step
)
key ∈ ri, key0 ∈ ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ ri[key0] = adr, key0 = key, adr ∈ fs, ri[key0] ∈ fs ∧ key-cons(key0, fs[ri[key0]])
User applied rule all left
And closed the goal!
Rule argument
pos: left 3 terms: key0 abort: F discard: T computed: T
Additional information
Used terms: key0