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Proof step 5 from proof pot-pot-02 (previous step)

0 ≤ j, 0 ≤ j0, ∀ j1, j, i. i→n(j1) < i→n(j0) → 0 ≤ j1 ∧ 0 ≤ j → i ^ (j * j1) = (i ^ j) ^ j1 ⊦ i ^ (j * j0) = (i ^ j) ^ j0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ 0 < i → j ^ i = j ^ i -1 * j
Proof step 7:
0 < j0, 0 ≤ j0, 0 ≤ j, ∀ j1, j, i. i→n(j1) < i→n(j0) → 0 ≤ j1 ∧ 0 ≤ j → i ^ (j * j1) = (i ^ j) ^ j1 ⊦ i ^ (j * j0) = (i ^ j) ^ j0 -1 * i ^ j

Rule argument

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

Additional information

Used simplifier rules: Pot-zero: ⊦ j ^ 0 = 1;% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s)% ⊦ i * 0 = 0; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: j0, i ^ j