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Proof step 8 from proof ri-cons-range-ri (
previous step
)
∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]), ∃ key. key ∈ ri ∧ ri[key] = adr ⊦ adr ∈ fs
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 9
:
key ∈ ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ ri[key] ∈ fs
Rule argument
Additional information