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Proof step 6 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 7:
⊦ j ≠ 0 → ~ i / j = ~(i / j)
Proof step 8:
2 ≠ 0, ~ 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-div inst: name: Div-left seq: ⊦ j ≠ 0 → ~ i / j = ~(i / j) subst: [j, i] → [2, abs(i) * 2 ^ j] flags: rewrite

Additional information

Used terms: 2, abs(i) * 2 ^ j