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Proof step 2 from proof append-mod (previous step)

n < # x + # x0, # x ≤ n ⊦ # x + # x0 = 0, (x + x0)[n, a] = x + x1 ↔ x0[n - # x, a] = x1, x0 = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
n = n0 + # x, n < # x + # x0, # x ≤ n ⊦ # x + # x0 = 0, (x + x0)[n, a] = x + x1 ↔ x0[n0, a] = x1, x0 = []

Rule argument

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

Additional information

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