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Proof step 41 from proof int2bytes-first-zerobyte (previous step)

0 < i, 0 ≤ i, 256 ≤ i, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 → (int2bytes(i0).first = i2b(0) ↔ n→i(# int2bits(i0)) % 8 = 1) ⊦ int2bytes(i / 256).first = i2b(0) ↔ (n→i(# int2bits(i / 256)) + n→i(# fillsgn(9, int2bits(i % 256)).rest)) % 8 = 1, n→i(# fillsgn(9, int2bits(i % 256)).rest) % 8 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ bits ≠ [] → # bits.rest = (# bits)-1 subst: [bits] → [fillsgn(9, int2bits(i % 256))] rotate: F spec: list-data inst: bitlista name: length-rest

Additional information

Used simplifier rules: ⊦ fillsgn(n, bits) = [] ↔ bits = [] ∧ n = 0;% ⊦ int2bits(j) ≠ []; Used terms: fillsgn(9, int2bits(i % 256))