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Proof step 12 from proof mod-2pot (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ 0 ≤ i ∧ 0 < j → (i * 2 ^ j) % 2 = 0

Rule argument

spec: <toplevel> name: mod2-2pot-03 seq: ⊦ 0 ≤ i ∧ 0 < j → (i * 2 ^ j) % 2 = 0 subst: [j, i] → [i, abs(j)] flags: <all paths>

Additional information

Used simplifier rules: ⊦ ~ i = 0 ↔ i = 0; (s)% ⊦ ~ i < 0 ↔ 0 < i; (s)% ⊦ 0 < abs(i) ↔ i ≠ 0; (s)% ⊦ 0 ≤ abs(i); (s) Used terms: i, abs(j)