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Proof step 26 from proof geom-mod (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 27:
⊦ ~ i - ~ j = j - i
Proof step 28:
~(i - 1) ≠ 0, ~(i - 1) ≠ 0, 1 - i ^ i0 = ~((i - 1) * j0), 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 + 1) → 0 ≤ j0 → (1 - i ^ j0) % (1 - i) = 0 ∨ i = 1 ⊦ i = 1, ~1 = i0, (~(i - 1) - ~(~((i - 1) * j0) * i)) % (i - 1) = 0

Rule argument

spec: int-abs inst: name: sub-minus seq: ⊦ ~ i - ~ j = j - i subst: [i, j] → [i - 1, ~((i - 1) * j0) * i] flags: rewrite

Additional information

Used terms: i - 1, ~((i - 1) * j0) * i