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Proof step 9 from proof geom-mod (previous step)

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

User applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
(1 - i ^ i0) % (1 - i) = 0, j = i0 + 1, 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, j = 0

Rule argument

spec: int-abs name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [j, 1, i0]

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: j, 1, i0