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Proof step 23 from proof bits2bytes-rec-02 (previous step)

n % 8 = 0, # bits = n + 8, 8 < n, ∀ bits0. # bits0 < n + 8 → true → # bits0 % 8 = 0 ∧ 8 < # bits0 → bits2bytes(bits0) = bits2bytes(firstn(8, bits0)) + bits2bytes(restn(8, bits0)) ⊦ bits2bytes(firstn(n, bits)) + i2b(bits2int(lastn(8, bits))) ' = bits2bytes(firstn(8, bits)) + bits2bytes(firstn(n - 8, restn(8, bits))) + i2b(bits2int(lastn(8, restn(8, bits)))) ', bits = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 24:
n = n0 + 8 ∧ 8 ≤ n, n % 8 = 0, # bits = n + 8, 8 < n, ∀ bits0. # bits0 < n + 8 → true → # bits0 % 8 = 0 ∧ 8 < # bits0 → bits2bytes(bits0) = bits2bytes(firstn(8, bits0)) + bits2bytes(restn(8, bits0)) ⊦ bits2bytes(firstn(n, bits)) + i2b(bits2int(lastn(8, bits))) ' = bits2bytes(firstn(8, bits)) + bits2bytes(firstn(n0, restn(8, bits))) + i2b(bits2int(lastn(8, restn(8, bits)))) ', bits = []

Rule argument

seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n, 8, 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, 8, n0