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Proof step 30 from proof shiftright-neg-div (previous step)

i + 2 ^ (abs(j) - 1) < 0, i < 0, abs(j) = 1 ∨ (i - 1) / 2 >> (abs(j) - 1) = ((i - 1) / 2 + 1 - 2 ^ (abs(j) - 1)) / 2 ^ (abs(j) - 1), ∀ i, j0. i→n(j0) < i→n(j) → i + 2 ^ (abs(j0) - 1) < 0 → j0 = 0 ∨ i >> abs(j0) = (i + 1 - 2 ^ abs(j0)) / 2 ^ abs(j0) ⊦ j = 0, (i - 1) / 2 >> (abs(j) - 1) = (i + 1 - 2 ^ abs(j)) / 2 ^ abs(j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 31:
⊦ 0 < i → j ^ i = j ^ i -1 * j
Proof step 32:
0 < abs(j), i + 2 ^ (abs(j) - 1) < 0, i < 0, abs(j) = 1 ∨ (i - 1) / 2 >> (abs(j) - 1) = ((i - 1) / 2 + 1 - 2 ^ (abs(j) - 1)) / 2 ^ (abs(j) - 1), ∀ i, j0. i→n(j0) < i→n(j) → i + 2 ^ (abs(j0) - 1) < 0 → j0 = 0 ∨ i >> abs(j0) = (i + 1 - 2 ^ abs(j0)) / 2 ^ abs(j0) ⊦ j = 0, (i - 1) / 2 >> (abs(j) - 1) = (i + 1 - 2 ^ abs(j) -1 * 2) / (2 ^ abs(j) -1 * 2)

Rule argument

seq: ⊦ 0 < i → j ^ i = j ^ i -1 * j subst: [i, j] → [abs(j), 2] rotate: F spec: int-pot inst: name: Pot-pos

Additional information

Used simplifier rules: ⊦ 0 < abs(i) ↔ i ≠ 0; Used terms: abs(j), 2