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Proof step 19 from proof bytes2int-zero-06 (previous step)

∀ by. by ∈ bytes → b2i(by) = 0, bytes2int(bytes.butlast) = 0 ∨ bytes = bytes.first ', ∀ bytes0. # bytes0 < # bytes → (∀ by. by ∈ bytes0 → b2i(by) = 0) → bytes2int(bytes0) = 0 ∨ bytes0 = [] ⊦ bytes = [], bytes2int(bytes.butlast) * 256 + band(b2i(bytes.last), 255) = 0, # bytes = 1

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 20:
bytes = by0 + bytes1, ∀ by. by ∈ bytes → b2i(by) = 0, bytes2int(bytes.butlast) = 0 ∨ bytes = by0 ', ∀ bytes0. # bytes0 < # bytes → (∀ by. by ∈ bytes0 → b2i(by) = 0) → bytes2int(bytes0) = 0 ∨ bytes0 = [] ⊦ bytes = [], bytes2int(bytes.butlast) * 256 + band(b2i(bytes.last), 255) = 0, # bytes = 1

Rule argument

seq: bytes ≠ [] ⊦ by = bytes.first ∧ bytes0 = bytes.rest ↔ bytes = by + bytes0 subst: [bytes, by, bytes0] → [bytes, by0, bytes1] rotate: T spec: list-data inst: bytelist name: elim-list-c

Additional information

Used simplifier rules: bytes ≠ [] ⊦ by = bytes.first ∧ bytes0 = bytes.rest ↔ bytes = by + bytes0; (elim) Used terms: bytes, by0, bytes1