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Proof step 3 from proof key-cons-ext (previous step)

key-cons(ndx0) ↔ (∀ n. n < # ndx0 → key-cons(ndx0[n].key, ndx0[n])) ⊦ key-cons(nd.key, nd) ∧ key-cons(ndx0) ↔ (∀ n. ¬ # ndx0 < n → key-cons((nd ' + ndx0)[n].key, (nd ' + ndx0)[n]))

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

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