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Proof step 2 from proof mod-two-03 (previous step)

0 < i ⊦ (i * 2 - 1) % 2 = 1

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-div inst: name: modeq-pos-04 seq: 0 ≤ i0 * j - j0, 0 < j, 0 < j0, j0 % j ≠ 0 ⊦ (i0 * j - j0) % j = j - j0 % j subst: [j0, j, i0] → [1, 2, i] flags: rewrite, current

Additional information

Used simplifier rules: sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ 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%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%zero-gr: ⊦ 0 < i → (0 < i * j ↔ 0 < j); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: 1, 2, i