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Proof step 4 from proof int2bytes-byte (previous step)

~128 ≤ i, i < 128 ⊦ bits2bytes(int2bits(i)) = i2b(i) '

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ # bits ≤ 8 → bits2bytes(bits) = i2b(bits2int(bits)) '
Proof step 6:
¬ # int2bits(i) ≤ 8, ~128 ≤ i, i < 128 ⊦ bits2bytes(int2bits(i)) = i2b(i) '

Rule argument

spec: bytes inst: name: bits2bytes-base seq: ⊦ # bits ≤ 8 → bits2bytes(bits) = i2b(bits2int(bits)) ' subst: [bits] → [int2bits(i)] flags: rewrite, current

Additional information

Used simplifier rules: int2int-inj: ⊦ bits2int(int2bits(i)) = i; from specification int2bits%c-04: ⊦ by ' = by0 ' ↔ by = by0; from specification list[bytelist] Used terms: int2bits(i)