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Proof step 6 from proof ri-cons-copy-new (previous step)

# kx = # ax, 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), (ri[kx, ax][key] ∈ fs ∨ ri[kx, ax][key] ∈ ax) ∧ key-cons(key, fs[ax, fs[ri[kx]]][ri[kx, ax][key]])

User applied rule cut formula

And obtained the following new goals

Proof step 7:
# kx = # ax, new(ax, fs), key ∈ ri, key ∈ kx, kx.toSet ⊆ dom ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ dups(ax), dups(kx), (ri[kx, ax][key] ∈ fs ∨ ri[kx, ax][key] ∈ ax) ∧ key-cons(key, fs[ax, fs[ri[kx]]][ri[kx, ax][key]])
Proof step 23:
# 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 ∈ kx, (ri[kx, ax][key] ∈ fs ∨ ri[kx, ax][key] ∈ ax) ∧ key-cons(key, fs[ax, fs[ri[kx]]][ri[kx, ax][key]])

Rule argument

fma: key ∈ kx

Additional information

Used formulas: key ∈ kx