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Proof step 7 from proof mod-twoodd-03 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: mod-twoodd seq: 0 ≤ i, 0 < j ⊦ (2 * i + 1) % (2 * j) = i % j * 2 + 1 subst: [j, i] → [j, ~ i - 1] flags: <all paths>

Additional information

Used simplifier rules: mod-neg-01: j ≠ 0 ⊦ ~ i % j = ~(i % j);%nummult-eq-01: j ≠ 0, j0 % j = 0 ⊦ i * (* j) = i0 * (* j0) ↔ i = i0 * j0 / j;% ⊦ ~ i - j = ~(i + j); (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ ~ i * j = ~(i * j); (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) = j - i; (s)% ⊦ i + (j - i) = j; (s)% ⊦ 1 * i = i; (s)% ⊦ 2 * i = 0 ↔ i = 0; (s)% ⊦ i - j = i0 ↔ i = i0 + j; (s)% ⊦ i ≤ ~ j ↔ i + j ≤ 0; (s)% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j; (s)% ⊦ i ≤ ~1 ↔ i < 0; (s) Used terms: j, ~ i - 1