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Proof step 20 from proof shiftright-shift (previous step)

i < 0 ∧ j ≠ 0, 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) ⊦ ((i - 1) / 2 >> (abs(j) - 1)) >> j0 = (i - 1) / 2 >> (j + j0 - 1), j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
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)

Rule argument

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ 0 < j → abs(j) = j;