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Proof step 2 from proof int2bytes-empty (previous step)

bits2bytes(int2bits(i)) = [] ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ 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-base: ⊦ # bits ≤ 8 → bits2bytes(bits) = i2b(bits2int(bits)) ';% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ bytes + by = bytes + by ';% ⊦ ¬ bytes + bytes0 ≠ [] ↔ bytes = [] ∧ bytes0 = [];% ⊦ by ' ≠ [];% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1; Used terms: int2bits(i)