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Proof step 3 from proof bits2bits-in (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ 8 < # bits → bits2bytes(bits) = bits2bytes(firstn(# bits - 8, bits)) + i2b(bits2int(lastn(8, bits)))
Proof step 5:
¬ 8 < # bits, ∀ boolvar, bits0. # bits0 < # bits → true → bits0 ≠ [] → (boolvar ∈ bytes2bits(bits2bytes(bits0)) ↔ boolvar ∈ bits0) ⊦ bits = [], boolvar ∈ bytes2bits(bits2bytes(bits)) ↔ boolvar ∈ bits
Proof step 10:
8 < # bits, ∀ boolvar, bits0. # bits0 < # bits → true → bits0 ≠ [] → (boolvar ∈ bytes2bits(bits2bytes(bits0)) ↔ boolvar ∈ bits0) ⊦ bits = [], boolvar ∈ bytes2bits(bits2bytes(firstn(# bits - 8, bits)) + i2b(bits2int(lastn(8, bits)))) ↔ boolvar ∈ bits

Rule argument

name: bits2bytes-rec rotate: F subst: [bits] → [bits]

Additional information

Used terms: bits