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

# ax1 = # ndx, new(ax1, fs), ri-cons(ri, fs) ⊦ ri-cons(ri, fs[ax1, ndx])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# ax1 = # ndx, new(ax1, fs), key ∈ ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ (ri[key] ∈ fs ∨ ri[key] ∈ ax1) ∧ key-cons(key, fs[ax1, ndx][ri[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)