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Proof step 3 from proof ri-cons-copy (previous step)

# 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]])

User applied rule all left

And obtained the following new goals

Proof step 4:
# ax1 = # ndx, new(ax1, fs), key ∈ ri, 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

pos: left 4 terms: key abort: F discard: T computed: T

Additional information

Used terms: key