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Proof step 2 from proof restn-firstn (previous step)

m ≤ n, m + n ≤ # x ⊦ restn(m, firstn(n, x)) = firstn(n - m, restn(m, x))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
n = n0 + m, m ≤ n, m + n ≤ # x ⊦ restn(m, firstn(n, x)) = firstn(n0, restn(m, x))

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n, m, n0] flags:

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, m, n0