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Proof step 2 from proof ri-cons-copy-new (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# ax = # kx, new(ax, fs), key ∈ ri[kx, ax], kx.toSet ⊆ dom ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ dups(kx), dups(ax), (ri[kx, ax][key] ∈ fs ∨ ri[kx, ax][key] ∈ ax) ∧ key-cons(key, fs[ax, fs[ri[kx]]][ri[kx, ax][key]])

Rule argument

Additional information

Used simplifier rules: ri-cons: ⊦ ri-cons(ri, fs) ↔ (∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]));%# ax = # ndx ⊦ adr ∈ fs[ax, ndx] ↔ adr ∈ fs ∨ adr ∈ ax; (s)% ⊦ # ri[kx] = # kx; (s)% ⊦ # fs[ax] = # ax; (s)