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Proof step 10 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(restn(8, bits))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# bits = n + 8, n % 8 = 0, n ≤ 8, ∀ 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(restn(8, bits)), bits = [], n = 0

Rule argument

Additional information

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