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Proof step 9 from proof div2-2pot-neg (previous step)

0 < j, ~ abs(i) < 0 ⊦ 2 = 0, ~((abs(i) * 2 ^ j) / 2) = ~ abs(i) * 2 ^ (j - 1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ 0 ≤ i ∧ 0 < j → (i * 2 ^ j) / 2 = i * 2 ^ (j - 1)

Rule argument

spec: <toplevel> name: div2-2pot-02 seq: ⊦ 0 ≤ i ∧ 0 < j → (i * 2 ^ j) / 2 = i * 2 ^ (j - 1) subst: [j, i] → [j, abs(i)] flags: <all paths>

Additional information

Used simplifier rules: ⊦ ~ i * j = ~(i * j); (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ ~ i = ~ j ↔ i = j; (s)% ⊦ ~ i < 0 ↔ 0 < i; (s)% ⊦ 0 < abs(i) ↔ i ≠ 0; (s)% ⊦ 0 ≤ abs(i); (s) Used terms: j, abs(i)