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Proof step 6 from proof geom-mod (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 7:
(1 - i ^ (j - 1)) % (1 - i) = 0 ∨ i = 1, 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

Rule argument

subst: [j0, i] → [j - 1, i]

Additional information

Used simplifier rules: sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%abs-pred: ⊦ i→n(i - 1) < i→n(i) ↔ 0 < i; from specification intnat Used terms: j - 1, i