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Proof step 4 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])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
key-cons(ndx0), key-cons(nd.key, nd), n ≤ # ndx0, ∀ n. n < # ndx0 → key-cons(ndx0[n].key, ndx0[n]) ⊦ key-cons((nd ' + ndx0)[n].key, (nd ' + ndx0)[n])

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)