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Proof step 6 from proof bits2bytes-rec-02 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
# bits = n + 8 ∧ 8 ≤ # bits, # bits % 8 = 0, 8 < # bits, ∀ bits0. # bits0 < # bits → true → # bits0 % 8 = 0 ∧ 8 < # bits0 → bits2bytes(bits0) = bits2bytes(firstn(8, bits0)) + bits2bytes(restn(8, bits0)) ⊦ bits = [], bits2bytes(firstn(n, bits)) + i2b(bits2int(lastn(8, bits))) ' = bits2bytes(firstn(8, bits)) + bits2bytes(restn(8, bits))

Rule argument

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