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Proof step 3 from proof div2-2pot-neg (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 4:
i = ~ abs(i), i < 0, 0 < j ⊦ (i * 2 ^ j) / 2 = i * 2 ^ (j - 1)

Rule argument

fma: i = ~ abs(i)

Additional information

Used simplifier rules: ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ ~ ~ i = i; (s) Used formulas: i = ~ abs(i)