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Proof step 2 from proof geom-mod (previous step)

0 ≤ j ⊦ (1 - i ^ j) % (1 - i) = 0, i = 1

User applied rule induction

And obtained the following new goals

Proof step 3:
0 ≤ j, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ j0 → (1 - i ^ j0) % (1 - i) = 0 ∨ i = 1 ⊦ (1 - i ^ j) % (1 - i) = 0 ∨ i = 1

Rule argument

indvar: i→n(j) wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: i→n(j)