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Proof step 31 from proof ri-cons-copy-new (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 32:
# kx = # ax, new(ax, fs), key ∈ ri, kx.toSet ⊆ dom ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ dups(ax), dups(kx), key-cons(key, fs[ri[key]]), ri[key] ∈ ax, key ∈ kx

Rule argument

Additional information

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