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Proof step 22 from proof geom-mod (previous step)

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

User applied rule insert equation

And obtained the following new goals

Proof step 23:
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 ⊦ ((j0 * ~ ~(1 - i)) * i - (i - 1)) % (i - 1) = 0, ~1 = i0, i = 1, ~ ~(1 - i) = 0

Rule argument

pos: left 2 flags: discard

Proof step 22 from proof geom-mod (previous step)

User applied rule insert equation

And obtained the following new goals

Rule argument

Additional information