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Proof step 2 from proof bits2int-2pot (previous step)

⊦ 2 ^ n→i(# bits) + bytes2int(bits2bytes(true ' + bits)) = 0, true ∈ bits

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ bytes2int(bytes) = bits2int(bytes2bits(bytes))
Proof step 4:
⊦ 2 ^ n→i(# bits) + bits2int(bytes2bits(bits2bytes(true ' + bits))) = 0, true ∈ bits

Rule argument

name: bytes2int rotate: F subst: [bytes] → [bits2bytes(true ' + bits)]

Additional information

Used terms: bits2bytes(true ' + bits)