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Proof step 9 from proof minus-odd (previous step)

0 ≤ j0 ⊦ i ^ (2 * j0 + 1) = i * i ^ (2 * j0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ 0 < i → j ^ i = j ^ i -1 * j

Rule argument

spec: <toplevel> name: Pot-pos seq: ⊦ 0 < i → j ^ i = j ^ i -1 * j subst: [i, j] → [2 * j0 + 1, i] flags: <all paths>

Additional information

Used simplifier rules: ⊦ (i + 1)-1 = i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)% ⊦ ~1 < i ↔ 0 ≤ i; (s)% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i ≤ j → (j < i ↔ false); (as)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult Used terms: 2 * j0 + 1, i