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Proof step 29 from proof bytes2int-append (previous step)

8 * # bytes0 = 0, bytes2bits(bytes).first ⊦ bitsinvert(bytes2bits(bytes)) + bitsinvert(bytes2bits(bytes0)) = [], (bits2uint(bitsinvert(bytes2bits(bytes)) + bitsinvert(bytes2bits(bytes0))) + 1) + band(bits2int(bytes2bits(bytes0)), 256 ^ n→i(# bytes0) - 1) = (bits2uint(bitsinvert(bytes2bits(bytes))) + 1) * 256 ^ n→i(# bytes0), bytes = [], bytes0 = [], bytes2bits(bytes) + bytes2bits(bytes0) = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 30:
⊦ m ≠ 0 → (m * n = 0 ↔ n = 0)

Rule argument

seq: ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0) subst: [n, m] → [# bytes0, 8] rotate: F spec: nat-mult inst: name: zero-06

Additional information

Used simplifier rules: ⊦ # bytes = 0 ↔ bytes = []; Used terms: # bytes0, 8