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Proof step 12 from proof geom-mod (previous step)

(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 - (i - 1)) % (1 - i) = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j)
Proof step 14:
1 - i ≠ 0, ∃ j0. 1 - i ^ i0 = j0 * (1 - i), 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 - (i - 1)) % (1 - i) = 0

Rule argument

spec: int-div inst: name: modex seq: j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j) subst: [j, i] → [1 - i, 1 - i ^ i0] flags: rewrite

Additional information

Used simplifier rules: sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%add-zero: ⊦ i + 0 = i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%sub-eq-01: ⊦ i = j → i - j = 0; from specification int-abs Used terms: 1 - i, 1 - i ^ i0