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Proof step 27 from proof shiftright-neg-div (previous step)

abs(j) - 1 = 0 ∨ (i - 1) / 2 >> abs(abs(j) - 1) = ((i - 1) / 2 + 1 - 2 ^ abs(abs(j) - 1)) / 2 ^ abs(abs(j) - 1), i + 2 ^ (abs(j) - 1) < 0, i < 0, ∀ 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 28:
⊦ ¬ j < 0 → abs(j) = j
Proof step 29:
¬ abs(j) - 1 < 0, abs(j) - 1 = 0 ∨ (i - 1) / 2 >> (abs(j) - 1) = ((i - 1) / 2 + 1 - 2 ^ (abs(j) - 1)) / 2 ^ (abs(j) - 1), i + 2 ^ (abs(j) - 1) < 0, i < 0, ∀ 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)

Rule argument

seq: ⊦ ¬ j < 0 → abs(j) = j subst: [j] → [abs(j) - 1] rotate: F spec: int-abs inst: name: Abs-pos

Additional information

Used simplifier rules: ⊦ abs(i) - 1 < 0 ↔ i = 0; Used terms: abs(j) - 1