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Proof step 2 from proof rest-sub (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

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

Rule argument

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

Additional information

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