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Proof step 6 from proof bytes2int-mklist-zeros-01 (previous step)

¬ 1 < # bytes, 0 ≤ b2i(bytes.first), ∀ n, bytes0. # bytes0 < # bytes → 0 ≤ b2i(bytes0.first) → bytes2int(mklist(i2b(0), n) + bytes0) = bytes2int(bytes0) ∨ bytes0 = [] ⊦ bytes = [], bytes2int(mklist(i2b(0), n) + bytes) = bytes2int(bytes)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# bytes = 1, 0 ≤ b2i(bytes.first), ∀ n, bytes0. # bytes0 < 1 → 0 ≤ b2i(bytes0.first) → bytes2int(mklist(i2b(0), n) + bytes0) = bytes2int(bytes0) ∨ bytes0 = [] ⊦ bytes = [], bytes2int(mklist(i2b(0), n) + bytes) = bytes2int(bytes)

Rule argument

Additional information

Used simplifier rules: ⊦ # bytes = 0 ↔ bytes = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);