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Proof step 8 from proof bytes2int-zero-03 (previous step)

bytes = bytes1 + by0 ', bytes2int(bytes1) * 256 + band(b2i(by0), 255) = 0, by ∈ bytes, ∀ by, bytes0. # bytes0 < # bytes → bytes2int(bytes0) = 0 ∧ by ∈ bytes0 → by = i2b(0) ⊦ by = i2b(0), # bytes = 1, bytes = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
bytes2int(bytes1) * 256 + band(b2i(by0), 255) = 0, ∀ by, bytes0. # bytes0 < #(bytes1 + by0 ') → bytes2int(bytes0) = 0 ∧ by ∈ bytes0 → by = i2b(0) ⊦ by = i2b(0), bytes1 = [], ¬ by ∈ bytes1 ∧ by ≠ by0

Rule argument

Additional information

Used simplifier rules: ⊦ by ∈ by0 ' ↔ by = by0;% ⊦ by ∈ bytes + bytes0 ↔ ¬ ¬(by ∈ bytes ∨ by ∈ bytes0);% ⊦ bytes + by ' ≠ [];% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ # by ' = 0 +1;% ⊦ #(bytes + bytes0) = # bytes + # bytes0;