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

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

User applied rule insert equation

And obtained the following new goals

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

Rule argument

pos: left 1 flags: discard

Additional information