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Proof step 10 from proof pot-dist-pos (previous step)

j = i0 +1, 0 < j + j0, 0 < j, 0 ≤ j, 0 ≤ j0, ∀ j1, j0, i. i→n(j1) < i→n(j) → 0 ≤ j1 ∧ 0 ≤ j0 → i ^ (j1 + j0) = i ^ j1 * i ^ j0 ⊦ i ^ (j + j0)-1 * i = (i ^ i0 * i) * i ^ j0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
0 ≤ j0, 0 ≤ i0 + j0, 0 ≤ i0, 0 ≤ i0 + 1, ∀ j1, j0, i. i→n(j1) < i→n(i0 +1) → 0 ≤ j1 ∧ 0 ≤ j0 → i ^ (j1 + j0) = i ^ j1 * i ^ j0 ⊦ i ^ ((i0 + 1) + j0)-1 * i = (i ^ i0 * i) * i ^ j0

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%succ-01: ⊦ i +1 = i + 1; from specification int-basic2