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Proof step 9 from proof ri-cons-subseteq (
previous step
)
key-cons(key, fs0[ri[key]]), ri[key] ∈ fs0, key ∈ ri, ri[key] ∈ fs1, ∀ adr. adr ∈ fs0 → adr ∈ fs1 ∧ fs0[adr] = fs1[adr] ⊦ key-cons(key, fs1[ri[key]])
User applied rule all left
And closed the goal!
Rule argument
pos: left 5 terms: ri[key] abort: F discard: T computed: T
Additional information
Used terms: ri[key]