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Proof step 50 from proof int2bytes-first-zerobyte (previous step)

n→i(# fillsgn(9, int2bits(i % 256)).rest) % 8 = 0, 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)) % 8 = 1

User applied rule apply induction

And closed the goal!

Rule argument

subst: [i0] → [i / 256]

Additional information

Used simplifier rules: ⊦ 0 < j ∧ 0 ≤ i → 0 ≤ i / j;% ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j);%j ≠ 0 ⊦ abs(i / j) = abs(i) / abs(j);% ⊦ 0 < j → abs(j) = j;% ⊦ 0 < j ∧ 0 ≤ i → (i / (* j) < j0 ↔ i < j * j0);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ 0 < i → (i * j ≤ i ↔ j ≤ 1);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i; Used terms: i / 256