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Proof step 6 from proof mod-twoodd-03 (previous step)

2 * j ≠ 0, 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

Rule argument

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

Additional information

Used simplifier rules: mod-neg-01: j ≠ 0 ⊦ ~ i % j = ~(i % j);%subsub: ⊦ ~ i - j = ~(i + j); from specification int-abs%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%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%Sub-def: ⊦ i + (j - i) = j; from specification int-abs%eq-pos: ⊦ 0 < i → (i * j = i * j0 ↔ j = j0); from specification int-mult%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%zero-neq-01: ⊦ 0 < i → (i * j = 0 ↔ j = 0); from specification int-mult%negle-07: ⊦ i ≤ ~ j ↔ i + j ≤ 0; from specification int-abs%numle-03: ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j; from specification int-abs%mone-05: ⊦ i ≤ ~1 ↔ i < 0; from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs Used terms: j, ~ i - 1