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Proof step 3 from proof keys-firstn (previous step)

n ≤ # ax ⊦ firstn(n, keys(fs, ax)) = keys(fs, firstn(n, ax))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
# ax0 = n, ax = ax0 + ax1, n ≤ # ax ⊦ firstn(n, keys(fs, ax)) = keys(fs, ax0)

Rule argument

spec: list-del, inst: address-list-set name: elim-firstn-restn seq: ⊦ n ≤ # ax → ( ax0 = firstn(n, ax) ∧ ax1 = restn(n, ax) ↔ ax = ax0 + ax1 ∧ # ax0 = n) subst: [ax, n, ax0, ax1] → [ax, n, ax0, ax1]

Additional information

Used simplifier rules: ⊦ n ≤ # ax → ( ax0 = firstn(n, ax) ∧ ax1 = restn(n, ax) ↔ ax = ax0 + ax1 ∧ # ax0 = n); (elim) Used terms: ax, n, ax0, ax1