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Proof step 4 from proof div2-2pot-neg (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ ~ i * j = ~(i * j)
Proof step 6:
~ abs(i) < j, ~ abs(i) < 0, 0 < j, ~ abs(i) + 1 < j ⊦ ~(abs(i) * 2 ^ j) / 2 = ~ abs(i) * 2 ^ (j - 1)

Rule argument

spec: int-mult inst: name: minus-left seq: ⊦ ~ i * j = ~(i * j) subst: [j, i] → [2 ^ j, abs(i)] flags: rewrite

Additional information

Used terms: 2 ^ j, abs(i)