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Proof step 17 from proof bits2bytes-len (previous step)

# bits ≤ 8, ∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ # bits2bytes(bits0) = (# bits0 - 1) / 8 + 1 ⊦ # i2b(bits2int(bits)) ' = (# bits - 1) / 8 + 1, bits = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
# bits ≤ 8, ∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ # bits2bytes(bits0) = (# bits0 - 1) / 8 + 1 ⊦ bits = [], (# bits - 1) / 8 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ # by ' = 0 +1;