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Proof step 140 from proof bytes2int-int2bytes-len-eq (previous step)

bytes2int(bytes1) * 256 + band(b2i(by), 255) = i + 256, bytes1 = bytes1.first ', ~129 < i, i < 0, ∀ bytes0. # bytes0 < #(bytes1 + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ bytes1 + by ' = i2b(0) ' + i2b(i) ', bytes1 = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

seq: bytes ≠ [] ⊦ by = bytes.first ∧ bytes0 = bytes.rest ↔ bytes = by + bytes0 subst: [bytes, by, bytes0] → [bytes1, by0, bytes] 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: bytes1, by0, bytes