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Proof step 10 from proof bits2bytes-rec-01 (previous step)

# bits = m * 8 + n, n + 1 < m * 8 + n, n < 8, 8 < m * 8 + n, n < m * 8 + n ⊦ n = 0, bits2bytes(firstn(n, bits) + restn(n, bits)) = bits2bytes(firstn(n, bits)) + bits2bytes(restn(n, bits)), bits = [], m * 8 = 1, m = 0, (m * 8 + n - n) % 8 = 0 ∧ restn(n, bits) ≠ []

User applied rule simplifier

And obtained the following new goals

Proof step 11:
restn(n, bits) = [], # bits = m * 8 + n, 8 < # bits, n < 8 ⊦ m = 0, m * 8 = 1, bits = [], n = 0

Rule argument

Additional information

Used simplifier rules: bits2bytes-empty: ⊦ bits2bytes(bits) ≠ [];% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes = bytes + bytes0 ↔ bytes0 = [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;%n ≠ 0 ⊦ (m * n) % n = 0;% ⊦ m + n - m = n;% ⊦ m < m + n ↔ n ≠ 0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m * n = n * m;% ⊦ m0 * m * n = (m0 * m) * n;% ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ m + n < m + n0 ↔ n < n0;