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Proof step 20 from proof bytes2int-len-one-01 (previous step)

b2i(by) < 128, 0 ≤ b2i(by) ⊦ bytes0 = [], bytes0.first = i2b(0), true ∈ bytes2bits(bytes0)

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] → [bytes0, 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: bytes0, by0, bytes