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Proof step 21 from proof shiftright-shift (previous step)

0 < j, j + j0 < 32, i < 0, 0 ≤ j0, 0 < j + j0, ∀ j0, i, j1. i→n(j1) < i→n(j) → i < 0 ∧ 0 ≤ j0 ∧ 0 ≤ j1 ∧ j1 + j0 < 32 → (i >> j1) >> j0 = i >> (j1 + j0) ⊦ j = 0, ((i - 1) / 2 >> (j - 1)) >> j0 = (i - 1) / 2 >> (j + j0 - 1)

User applied rule apply induction

And closed the goal!

Rule argument

subst: [j0, i, j1] → [j0, (i - 1) / 2, j - 1]

Additional information

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