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Proof step 9 from proof key-cons-ext (previous step)

n = m + 1, key-cons(nd.key, nd), key-cons(ndx0), n ≤ # ndx0, ∀ n. n < # ndx0 → key-cons(ndx0[n].key, ndx0[n]) ⊦ n = 0, key-cons(ndx0[m].key, ndx0[m])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
m < # ndx0, key-cons(nd.key, nd), key-cons(ndx0), ∀ n. n < # ndx0 → key-cons(ndx0[n].key, ndx0[n]) ⊦ ndx0 = [], key-cons(ndx0[m].key, ndx0[m])

Rule argument

Additional information

Used simplifier rules: ⊦ # ndx = 0 ↔ ndx = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)