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Proof step 1 from proof lastn-restn-01 (conclusion)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 2:
# x = n0 + n, n ≤ # x ⊦ lastn(n, x) = restn(n0, 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