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Proof step 29 from proof bytes2int-diff (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 30:
# 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

Rule argument

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i * j = j * i;