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Proof step 16 from proof intshiftrsign-2pot-neg-01 (previous step)

0 < j, j < 32, 0 ≤ j, i0 < 0, ∀ i, j0. i→n(j0) < i→n(j) → true → i < 0 ∧ 0 < j0 ∧ j0 < 32 → i i>> j0 = (i - (2 ^ j0 - 1)) / 2 ^ j0 ⊦ j = 1, (i0 + 1 - 2 ^ (j - 1)) / 2 ^ (j - 1) = (2 * (i0 + 1) - 1 + 2 * 2 ^ (j - 1)) / (2 * 2 ^ (j - 1))

User applied rule cut formula

And obtained the following new goals

Proof step 17:
2 * (i0 + 1) - 1 + 2 * 2 ^ (j - 1) = 2 * (i0 + 1 - 2 ^ (j - 1)) - 1, 0 < j, j < 32, 0 ≤ j, i0 < 0, ∀ i, j0. i→n(j0) < i→n(j) → true → i < 0 ∧ 0 < j0 ∧ j0 < 32 → i i>> j0 = (i - (2 ^ j0 - 1)) / 2 ^ j0 ⊦ j = 1, (i0 + 1 - 2 ^ (j - 1)) / 2 ^ (j - 1) = (2 * (i0 + 1) - 1 + 2 * 2 ^ (j - 1)) / (2 * 2 ^ (j - 1))

Rule argument

fma: 2 * (i0 + 1) - 1 + 2 * 2 ^ (j - 1) = 2 * (i0 + 1 - 2 ^ (j - 1)) - 1

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i + (j - i) = j;% ⊦ i * j + i * j0 = i * (j + j0);% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ 0 < i → (i * j = i * j0 ↔ j = j0); Used formulas: 2 * (i0 + 1) - 1 + 2 * 2 ^ (j - 1) = 2 * (i0 + 1 - 2 ^ (j - 1)) - 1