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Proof step 3 from proof bytes2int-neg (previous step)

∀ bytes0. # bytes0 < # bytes → true → bytes0 ≠ [] → (bytes2int(bytes0) < 0 ↔ b2i(bytes0.first) < 0) ⊦ bytes = [], bytes2int(bytes) < 0 ↔ ¬ 0 ≤ b2i(bytes.first)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
bytes = by + bytes1, ∀ bytes0. # bytes0 < # bytes → true → bytes0 ≠ [] → (bytes2int(bytes0) < 0 ↔ b2i(bytes0.first) < 0) ⊦ bytes = [], bytes2int(bytes) < 0 ↔ ¬ 0 ≤ b2i(by)

Rule argument

seq: bytes ≠ [] ⊦ by = bytes.first ∧ bytes0 = bytes.rest ↔ bytes = by + bytes0 subst: [bytes, by, bytes0] → [bytes, by, 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, by, bytes1