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Proof step 21 from proof geom-mod (previous step)

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 - i = 0, ~1 = i0, i = 1

User applied rule cut formula

And obtained the following new goals

Proof step 22:
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

Rule argument

fma: 1 - i = ~ ~(1 - i)

Additional information

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