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Proof step 12 from proof uint2bits-len-mon (previous step)

0 < j, # uint2bits(j / 2) < # uint2bits(i / 2), i ≤ j, 0 ≤ i, 0 ≤ j, 0 < i, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ j0 ∧ # uint2bits(j0) < # uint2bits(i) ∧ i ≤ j0 ∧ 0 ≤ i → false ⊦ i = 1, j = 1

User applied rule apply induction

And closed the goal!

Rule argument

subst: [j0, i] → [j / 2, i / 2]

Additional information

Used simplifier rules: ⊦ 0 < i → (i * j ≤ i ↔ j ≤ 1);% ⊦ 0 < j → abs(j) = j;%j ≠ 0 ⊦ abs(i / j) = abs(i) / abs(j);% ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j);% ⊦ 0 ≤ i / 2 ↔ ~1 ≤ i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ 0 < j ∧ 0 ≤ i → (i / (* j) < j0 ↔ i < j * j0);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)% ⊦ m ≤ n ∧ n < n0 → m < n0; (forward)% ⊦ ¬ n < n;% ⊦ i < j ∧ j < i0 → i +1 < i0; (forward)% ⊦ i < j ∧ j < j0 → i < j0; (forward)% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ 0 < i ∧ 0 ≤ j → ((* j) < (* i) * i0 ↔ j / i < i0);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ 0 < i → (0 < i * j ↔ 0 < j);% ⊦ 0 ≤ i ∧ 0 < j → ¬ i < i / j * j; Used terms: j / 2, i / 2