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Proof step 9 from proof shiftright-shift (previous step)

0 ≤ j0, i < 0, j + j0 < 32, 0 < j, ∀ 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 >> j) >> j0 = i >> (j + j0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ i < 0 ∧ j ≠ 0 → i >> j = (i - 1) / 2 >> (abs(j) - 1)
Proof step 11:
¬(i < 0 ∧ j + j0 ≠ 0), 0 ≤ j0, i < 0, j + j0 < 32, 0 < j, ∀ 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 >> j) >> j0 = i >> (j + j0)
Proof step 14:
i < 0 ∧ j + j0 ≠ 0, 0 ≤ j0, i < 0, j + j0 < 32, 0 < j, ∀ 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 >> j) >> j0 = (i - 1) / 2 >> (abs(j + j0) - 1)

Rule argument

name: shiftright-rec-neg rotate: F subst: [j, i] → [j + j0, i]

Additional information

Used terms: j + j0, i