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Proof step 22 from proof bits2bytes-rec-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
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 = []

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0;%n0 < n ⊦ (* n) < m0 + (* n0) ↔ n - n0 < m0;% ⊦ 0 + m = m;% ⊦ n ≤ m → m + n0 - n = (m - n) + n0;% ⊦ m < n ∧ n = n0 → m < n0; (forward)%n0 ≤ n ⊦ m < n - n0 ↔ m + n0 < n;% ⊦ bytes + by = bytes + by ';%n ≤ # bits ⊦ # restn(n, bits) = # bits - n;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward)% ⊦ m < n ∧ m = n0 → n0 < n; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward)% ⊦ # bits = 0 ↔ bits = [];% ⊦ m +1 = m + 1;% ⊦ ¬ m + n < m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;