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Proof step 21 from proof ri-cons-copy-new (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
# 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

Rule argument

Additional information

Used simplifier rules: ⊦ fs[adr, nd][adr] = nd; (s)% ⊦ # ri[kx] = # kx; (s)% ⊦ # fs[ax] = # ax; (s)