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Proof step 8 from proof mod-2pot (previous step)

0 < i, ~ abs(j) < 0 ⊦ (~ abs(j) * 2 ^ i) % 2 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ ~ i * j = ~(i * j)
Proof step 10:
~ abs(j) < 0, 0 < i ⊦ ~(abs(j) * 2 ^ i) % 2 = 0

Rule argument

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

Additional information

Used terms: 2 ^ i, abs(j)