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Proof step 6 from proof bytes2int-zero-03 (previous step)

1 < # bytes, bytes2int(bytes.butlast) * 256 + band(b2i(bytes.last), 255) = 0, by ∈ bytes, ∀ by, bytes0. # bytes0 < # bytes → bytes2int(bytes0) = 0 ∧ by ∈ bytes0 → by = i2b(0) ⊦ by = i2b(0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
bytes2int(bytes.butlast) * 256 + band(b2i(bytes.last), 255) = 0, by ∈ bytes, ∀ by, bytes0. # bytes0 < # bytes → bytes2int(bytes0) = 0 ∧ by ∈ bytes0 → by = i2b(0) ⊦ by = i2b(0), # bytes = 1, bytes = []

Rule argument

Additional information

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