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Proof step 12 from proof bits2bytes-app (previous step)

# bits0 ≤ 8, # bits0 % 8 = 0, # bits + # bits0 ≤ 8, ∀ bits1, bits2. # bits2 + # bits1 < # bits + # bits0 → # bits1 % 8 = 0 → bits2 = [] ∨ bits1 = [] ∨ bits2bytes(bits2 + bits1) = bits2bytes(bits2) + bits2bytes(bits1) ⊦ i2b(bits2int(bits + bits0)) ' = bits2bytes(bits) + i2b(bits2int(bits0)) ', bits0 = [], bits = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ n ≠ 0 → (m % n = 0 ↔ (∃ n0. m = n0 * n))
Proof step 14:
8 ≠ 0, # bits0 ≤ 8, ∃ n0. # bits0 = n0 * 8, # bits + # bits0 ≤ 8, ∀ bits1, bits2. # bits2 + # bits1 < # bits + # bits0 → # bits1 % 8 = 0 → bits2 = [] ∨ bits1 = [] ∨ bits2bytes(bits2 + bits1) = bits2bytes(bits2) + bits2bytes(bits1) ⊦ i2b(bits2int(bits + bits0)) ' = bits2bytes(bits) + i2b(bits2int(bits0)) ', bits0 = [], bits = []

Rule argument

seq: ⊦ n ≠ 0 → (m % n = 0 ↔ (∃ n0. m = n0 * n)) subst: [n, m] → [8, # bits0] rotate: F spec: nat-div inst: name: modex

Additional information

Used terms: 8, # bits0