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Proof step 5 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

User applied rule cut formula

And obtained the following new goals

Proof step 6:
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

fma: j = 0

Additional information

Used simplifier rules: Pot-zero: ⊦ j ^ 0 = 1;% ⊦ i - j = i0 ↔ i = i0 + j; (s)% ⊦ i + 0 = i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%j ≠ 0 ⊦ 0 % j = 0; (s) Used formulas: j = 0