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Proof step 3 from proof minus-square (previous step)

0 ≤ j, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ j0 → ~ i ^ (2 * j0) = i ^ (2 * j0) ⊦ ~ i ^ (2 * j) = i ^ (2 * j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ 0 ≤ j → i ^ (2 * j) = (i * i) ^ j

Rule argument

spec: int-pot inst: name: pot-square seq: ⊦ 0 ≤ j → i ^ (2 * j) = (i * i) ^ j subst: [j, i] → [j, ~ i] flags: rewrite, current

Additional information

Used simplifier rules: pot-square-02: ⊦ 0 ≤ j → ((i * i) ^ j ≠ i ^ (2 * j) ↔ false);%minus-left: ⊦ ~ i * j = ~(i * j); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%neg-02: ⊦ ~ ~ i = i; from specification int-abs Used terms: j, ~ i