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Proof step 11 from proof shiftright-is-le (previous step)

i < i / 2 >> (abs(j) - 1), 0 ≤ i, 0 < i / 2 >> (abs(j) - 1), ∀ j, i0. i→n(i0) < i→n(i) → abs(i0) < i0 >> j → false ⊦ j = 0

User applied rule apply induction

And closed the goal!

Rule argument

subst: [j, i0] → [abs(j) - 1, i / 2]

Additional information

Used simplifier rules: shiftright-zero: ⊦ 0 >> j = 0;% ⊦ 0 < i → (i * j < i ↔ j < 1);% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j);%j ≠ 0 ⊦ abs(i / j) = abs(i) / abs(j);% ⊦ 0 ≤ i → abs(i) = i;% ⊦ 0 < j ∧ 0 ≤ i → (i / (* j) < j0 ↔ i < j * j0);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ (* i1) * j ≤ j ↔ 0 ≤ (1 - i1) * j;% ⊦ i * ~1 = ~ i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ i ≤ ~ j ↔ i + j ≤ 0;% ⊦ i + 0 = i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i ≤ j → (j ≤ i ↔ i = j); Used terms: abs(j) - 1, i / 2