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Proof step 22 from proof ri-cons-copy-new (previous step)

# kx0 = # ax, # ax1 = # kx1, new(ax, fs), new(ax1, fs), key ∈ ri, kx0.toSet ⊆ dom ri, kx1.toSet ⊆ dom ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ dups(ax), dups(ax1), dups(ax1 + ax), dups(kx1 + kx0), dups(kx1), dups(kx0), key-cons(key, fs[ri[key]]), adr ∈ ax, adr ∈ ax1, key ∈ kx1, key ∈ kx0, adr ∈ fs

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 8 terms: key

Additional information

Used terms: key