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Proof step 9 from proof uint2uint-pos (previous step)

i % 2 = 0, 0 < i, 0 ≤ i, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 → bits2uint(uint2bits(i0)) = i0 ⊦ i = 1, bits2uint(uint2bits(i / 2)) * 2 = i

User applied rule apply induction

And closed the goal!

Rule argument

subst: [i0] → [i / 2]

Additional information

Used simplifier rules: ⊦ i < j ∧ j = i0 → i < i0; (forward)% ⊦ i % 2 = 0 → (i / 2 = j ↔ i = 2 * j);% ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ 0 ≤ i / 2 ↔ ~1 ≤ i;% ⊦ 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 / 2