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Proof step 19 from proof bits2bytes-rec-02 (previous step)

# restn(8, bits) ≤ 8, # bits = 16, ∀ bits0. # bits0 < 16 → true → # bits0 % 8 = 0 ∧ 8 < # bits0 → bits2bytes(bits0) = bits2bytes(firstn(8, bits0)) + bits2bytes(restn(8, bits0)) ⊦ bits = [], i2b(bits2int(lastn(8, bits))) ' = i2b(bits2int(restn(8, bits))) '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
# bits = 16, ∀ bits0. # bits0 < 16 → true → # bits0 % 8 = 0 ∧ 8 < # bits0 → bits2bytes(bits0) = bits2bytes(firstn(8, bits0)) + bits2bytes(restn(8, bits0)) ⊦ bits2int(lastn(8, bits)) = bits2int(restn(8, bits)), i2b(bits2int(lastn(8, bits))) = i2b(bits2int(restn(8, bits))), bits = []

Rule argument

Additional information

Used simplifier rules: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)%n0 ≤ n ⊦ n - n0 ≤ m ↔ n ≤ m + n0;%n ≤ # bits ⊦ # restn(n, bits) = # bits - n;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < n ∧ m = n0 → n0 < n; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ i2b(i) ≠ i2b(j) → i ≠ j; (forward)% ⊦ by ' = by0 ' ↔ by = by0;