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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
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

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)