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Proof step 7 from proof key-cons-ext (previous step)

key-cons(nd.key, nd), key-cons(ndx0), n ≤ # ndx0, ∀ n. n < # ndx0 → key-cons(ndx0[n].key, ndx0[n]) ⊦ n < # nd ', key-cons(ndx0[n - # nd '].key, ndx0[n - # nd '])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
key-cons(ndx0), key-cons(nd.key, nd), n ≤ # ndx0, ∀ n. n < # ndx0 → key-cons(ndx0[n].key, ndx0[n]) ⊦ n = 0, key-cons(ndx0[n - 1].key, ndx0[n - 1])

Rule argument

Additional information

Used simplifier rules: ⊦ m < 1 ↔ m = 0; (s)% ⊦ # nd ' = 0 +1; (s)