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Proof step 36 from proof bytes2int-diff (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 37:
b2i(by0) < 128, 256 + b2i(by) = b2i(by0), # bytes2 = # bytes3, bytes2int(bytes2) = bytes2int(bytes3), 0 ≤ b2i(by0), b2i(by) < 0, ∀ bytes0, bytes1. # bytes1 < #(bytes3 + by0 ') → # bytes0 = # bytes1 ∧ bytes2int(bytes0) = bytes2int(bytes1) → bytes0 = bytes1 ⊦ by = by0, bytes2 + by ' = bytes3 + by0 ', bytes3 = []

Rule argument

fma: b2i(by0) < 128

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;%i ≥ 128 ⊦ ¬ (* i) ≤ b2i(by); Used formulas: b2i(by0) < 128