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Proof step 7 from proof int2bytes-minus-one (previous step)

bits2bytes(int2bits(i)) = i2b(~1) ', 8 < # int2bits(i) ⊦ i = ~1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ 8 < # bits → bits2bytes(bits) = bits2bytes(firstn(# bits - 8, bits)) + i2b(bits2int(lastn(8, bits)))

Rule argument

name: bits2bytes-rec rotate: F subst: [bits] → [int2bits(i)]

Additional information

Used simplifier rules: bits2bytes-empty: ⊦ bits2bytes(bits) ≠ [];% ⊦ bytes + by = bytes + by ';% ⊦ by ' = bytes + by0 ' ↔ by = by0 ∧ bytes = [];% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1; Used terms: int2bits(i)