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Proof step 15 from proof shiftright-shift (previous step)

0 < j, j + j0 < 32, i < 0, 0 ≤ 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 + j0 = 0, (i >> j) >> j0 = (i - 1) / 2 >> (abs(j + j0) - 1), j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ ¬ j < 0 → abs(j) = j
Proof step 17:
¬ j + j0 < 0, 0 < j, j + j0 < 32, i < 0, 0 ≤ 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 + j0 = 0, (i >> j) >> j0 = (i - 1) / 2 >> (j + j0 - 1), j = 0

Rule argument

seq: ⊦ ¬ j < 0 → abs(j) = j subst: [j] → [j + j0] rotate: F spec: int-abs inst: name: Abs-pos

Additional information

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