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Proof step 22 from proof bytes2int-diff (previous step)

bytes2int(bytes2) * 256 + band(b2i(by), 255) = bytes2int(bytes.butlast) * 256 + band(b2i(bytes.last), 255), # bytes2 + 1 = # bytes, ∀ bytes0, bytes1. # bytes1 < # bytes → # bytes0 = # bytes1 ∧ bytes2int(bytes0) = bytes2int(bytes1) → bytes0 = bytes1 ⊦ bytes = [], # bytes = 1, bytes2 + by ' = bytes

User applied rule insert elim lemma

And obtained the following new goals

Proof step 23:
bytes = bytes3 + by0 ', bytes2int(bytes2) * 256 + band(b2i(by), 255) = bytes2int(bytes3) * 256 + band(b2i(by0), 255), # bytes2 + 1 = # bytes, ∀ bytes0, bytes1. # bytes1 < # bytes → # bytes0 = # bytes1 ∧ bytes2int(bytes0) = bytes2int(bytes1) → bytes0 = bytes1 ⊦ bytes = [], # bytes = 1, bytes2 + by ' = bytes

Rule argument

seq: bytes ≠ [] ⊦ by = bytes.last ∧ bytes0 = bytes.butlast ↔ bytes = bytes0 + by ' subst: [bytes, by, bytes0] → [bytes, by0, bytes3] rotate: T spec: list-last inst: bytelist name: last-elim

Additional information

Used simplifier rules: bytes ≠ [] ⊦ by = bytes.last ∧ bytes0 = bytes.butlast ↔ bytes = bytes0 + by '; (elim) Used terms: bytes, by0, bytes3