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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
(1 - i ^ i0) % (1 - i) = 0, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → 0 ≤ j0 → (1 - i ^ j0) % (1 - i) = 0 ∨ i = 1 ⊦ i = 1, ~1 = i0, (1 - i ^ i0 * i) % (1 - i) = 0

Rule argument

Additional information

Used simplifier rules: succ: ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j;%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs