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Proof step 32 from proof bytes2int-diff (previous step)

¬ 0 ≤ b2i(by), # bytes2 = # bytes3, band(b2i(by), 255) = band(b2i(by0), 255), bytes2int(bytes2) = bytes2int(bytes3), ∀ bytes0, bytes1. # bytes1 < #(bytes3 + by0 ') → # bytes0 = # bytes1 ∧ bytes2int(bytes0) = bytes2int(bytes1) → bytes0 = bytes1 ⊦ bytes3 = [], bytes2 + by ' = bytes3 + by0 ', by = by0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 33:
bytes2int(bytes2) = bytes2int(bytes3), # bytes2 = # bytes3, 256 + b2i(by) = band(b2i(by0), 255), b2i(by) < 0, ∀ bytes0, bytes1. # bytes1 < #(bytes3 + by0 ') → # bytes0 = # bytes1 ∧ bytes2int(bytes0) = bytes2int(bytes1) → bytes0 = bytes1 ⊦ bytes3 = [], bytes2 + by ' = bytes3 + by0 ', by = by0

Rule argument

Additional information

Used simplifier rules: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ band(i, j) = band(j, i);% ⊦ band(i, band(i0, j)) = band(band(i, i0), j);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (b2i(by) + 1) % 256 = b2i(by) + 1;% ⊦ i < 0 → band(i, 255) = (i + 1) % 256 + 255;% ⊦ ¬ i ≤ j ↔ j < i;