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Proof step 4 from proof mod-twoodd-03 (previous step)

i < j, i < 0, 0 < j, i + 1 < j ⊦ (i + 1) % j * 2 = ~(2 * (~ i - 1) + 1) % (2 * j) + 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
j ≠ 0 ⊦ ~ i % j = ~(i % j)
Proof step 6:
2 * j ≠ 0, i < j, i < 0, 0 < j, i + 1 < j ⊦ (i + 1) % j * 2 = ~((2 * (~ i - 1) + 1) % (2 * j)) + 1

Rule argument

spec: int-div inst: name: mod-neg-01 seq: j ≠ 0 ⊦ ~ i % j = ~(i % j) subst: [j, i] → [2 * j, 2 * (~ i - 1) + 1] flags: rewrite, current

Additional information

Used simplifier rules: zero-neq-01: ⊦ 0 < i → (i * j = 0 ↔ j = 0); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%subsub: ⊦ ~ i - j = ~(i + j); from specification int-abs%minus-left: ⊦ ~ i * j = ~(i * j); from specification int-mult%neg-15: ⊦ i + ~ j = i - j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%minus-sub: ⊦ ~(i - j) = j - i; from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1 Used terms: 2 * j, 2 * (~ i - 1) + 1