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Proof step 33 from proof bytes2int-diff (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 34:
⊦ b2i(by) < 0 → band(b2i(by), 255) = b2i(by) + 256
Proof step 35:
¬ b2i(by0) < 0, 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

seq: ⊦ b2i(by) < 0 → band(b2i(by), 255) = b2i(by) + 256 subst: [by] → [by0] rotate: F spec: byte inst: name: byte-011x-neg

Additional information

Used simplifier rules: ⊦ i + j = i + i0 ↔ j = i0;% ⊦ b2i(byi) = b2i(byj) ↔ byi = byj;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i; Used terms: by0