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Proof step 10 from proof bytes2int-fillsgn-zero (previous step)

# bytes = 1, b2i(bytes.first) = 0 ⊦ bytes = [], fillsgn(1, i2b(0) ') = bytes

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
bytes = by + bytes0, # bytes = 1, b2i(by) = 0 ⊦ bytes = [], fillsgn(1, i2b(0) ') = bytes

Rule argument

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