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Proof step 11 from proof fill-lastn-03 (previous step)

m < n, # bits < n, m < # bits, m + 1 < n ⊦ lastn(m, mklist(bits.first, n - # bits) + bits) = lastn(m, bits), bits = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 12:
n = n0 + # bits ∧ # bits ≤ n, m < n, # bits < n, m < # bits, m + 1 < n ⊦ lastn(m, mklist(bits.first, n0) + bits) = lastn(m, bits), bits = []

Rule argument

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

Additional information

Used simplifier rules: ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < n → ¬ n < m;%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, # bits, n0