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Proof step 6 from proof bits2bits-in (previous step)

# bits ≤ 8, ∀ boolvar, bits0. # bits0 < # bits → true → bits0 ≠ [] → (boolvar ∈ bytes2bits(bits2bytes(bits0)) ↔ boolvar ∈ bits0) ⊦ bits = [], boolvar ∈ byte2bits(i2b(bits2int(bits))) ↔ boolvar ∈ bits

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ byte2bits(by) = fillsgn(8, int2bits(b2i(by)))
Proof step 8:
# bits ≤ 8, ∀ boolvar, bits0. # bits0 < # bits → true → bits0 ≠ [] → (boolvar ∈ bytes2bits(bits2bytes(bits0)) ↔ boolvar ∈ bits0) ⊦ bits = [], boolvar ∈ fillsgn(8, int2bits(b2i(i2b(bits2int(bits))))) ↔ boolvar ∈ bits

Rule argument

name: byte2bits rotate: F subst: [by] → [i2b(bits2int(bits))]

Additional information

Used terms: i2b(bits2int(bits))