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Proof step 5 from proof succ (previous step)

0 < i + 1, 0 ≤ i ⊦ j ^ (i + 1)-1 * j = j ^ i * j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
0 ≤ i ⊦ j ^ (i + 1)-1 * j = j ^ i * j

Rule argument

Additional information

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