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Proof step 24 from proof geom-mod (previous step)

~(i - 1) ≠ 0, ~(i - 1) ≠ 0, 1 - i ^ i0 = j0 * ~(i - 1), 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, ((j0 * ~(i - 1)) * i - (i - 1)) % (i - 1) = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 25:
⊦ ~ i * j = ~(i * j)
Proof step 26:
~(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

Rule argument

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

Additional information

Used simplifier rules: commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: j0, i - 1